Leetcode-做题学算法周刊第三期

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首发于微信公众号《前端成长记》，写于 2019.11.13

背景

本文记录刷题过程中的整个思考过程，以供参考。主要内容涵盖：

  • 题目分析设想
  • 编写代码验证
  • 查阅他人解法
  • 思考总结

目录

  • 35. 搜索插入位置
  • 38. 报数
  • 53. 最大子序和
  • 58. 最后一个单词的长度
  • 66. 加一

Easy

35. 搜索插入位置

题目地址

题目描述

给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。

你可以假设数组中无重复元素。

示例：

输入: [1,3,5,6], 5
输出: 2

输入: [1,3,5,6], 2
输出: 1

输入: [1,3,5,6], 7
输出: 4

输入: [1,3,5,6], 0
输出: 0

题目分析设想

这道题目有点明显，题干说明了是排序数组，重点是排序数组，所以很明显的第一反应会使用二分法来解题。同时可以注意一下，数组中无重复元素。所以这道题我就按两个方案来作答：

  • 暴力法，直接遍历
  • 二分法，可以理解成不断折半排除不可能

编写代码验证

Ⅰ. 暴力法

代码：

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var searchInsert = function(nums, target) {if (nums.length === 0 || nums[0] > target) return 0;
    if(nums[nums.length - 1] < target) return nums.length;

    for(let i = 0; i < nums.length; i++) {if(nums[i] >= target) return i
    }
};

结果：

  • 62/62 cases passed (60 ms)
  • Your runtime beats 92.48 % of javascript submissions
  • Your memory usage beats 63.22 % of javascript submissions (33.8 MB)
  • 时间复杂度 O(n)

Ⅱ. 二分法

代码：

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var searchInsert = function(nums, target) {if (nums.length === 0 || nums[0] > target) return 0;
    if(nums[nums.length - 1] < target) return nums.length;

    let left = 0; // 起点
    let right = nums.length - 1; // 终点
    while(left < right) {// 零填充右位移，使用位运算避免溢出，大部分情况下等于 (left + right / 2)
        let i = parseInt((left + right) >>> 1) // 这里选择取左
        if (nums[i] < target) { // 中位数小于目标值
            left = i + 1 // 排除中位数左侧
        } else {right = i // 排除中位数右侧}
    }
    return left
};

结果：

  • 62/62 cases passed (52 ms)
  • Your runtime beats 99.31 % of javascript submissions
  • Your memory usage beats 61.31 % of javascript submissions (33.8 MB)
  • 时间复杂度 O(log2(n))

查阅他人解法

基本上这道题就是针对二分法进行考察的，所以没有看到其他特别的解法。

思考总结

看见排序数组，查找下标，那么就可以果断选择二分法啦。

38. 报数

题目地址

题目描述

报数序列是一个整数序列，按照其中的整数的顺序进行报数，得到下一个数。其前五项如下：

1.     1
2.     11
3.     21
4.     1211
5.     111221

1 被读作 "one 1" ("一个一"), 即 11。

11 被读作 "two 1s" ("两个一"), 即 21。

21 被读作 "one 2", "one 1"（"一个二" , "一个一") , 即 1211。

给定一个正整数 n（1 ≤ n ≤ 30），输出报数序列的第 n 项。

注意：整数顺序将表示为一个字符串。

示例:

输入: 1
输出: "1"

输入: 4
输出: "1211"

题目分析设想

这道题目有点绕，其实我们只看右侧项就可以，每次都是读上一次项。报数的规则实际上就是相同连续数字合并后进行每位的报数。最简单的想法是直接使用正则替换就可以了。当然也可以从递归和遍历的方式来作答，我们分别来看看。

  • 正则法，替换相同连续数字为 长度 + 数字本身
  • 递归法，不断转换成 n-1 求解
  • 遍历法，不断转换成 n+1 求解

编写代码验证

Ⅰ. 正则法

代码：

/**
 * @param {number} n
 * @return {string}
 */
var countAndSay = function(n) {
    let str = '1'
    for(let i = 1; i < n; i++) {
        // 匹配项长度 + 第一位即为读数
        str = str.replace(/(\d)\1*/g, (match) => (`${match.length}${match.charAt(0)}`))
    }
    return str
};

结果：

  • 18/18 cases passed (56 ms)
  • Your runtime beats 98.81 % of javascript submissions
  • Your memory usage beats 32.53 % of javascript submissions (35.6 MB)
  • 时间复杂度 O(n)

Ⅱ. 递归法

代码：

/**
 * @param {number} n
 * @return {string}
 */
var countAndSay = function(n) {if (n === 1) return '1'
    debugger
    let str = countAndSay(n - 1)
    let item = str.charAt(0)

    let count = 0 // 计数器
    let res = ''
    for(let i = 0; i < str.length; i++) {if(str.charAt(i) === item) {count++} else {res += `${count}${item}`
            item = str.charAt(i)
            count = 1
        }

        if (i === str.length - 1) { // 最后一位，需要取数
            res += `${count}${item}`
        }
    }
    return res
};

结果：

  • 18/18 cases passed (64 ms)
  • Your runtime beats 92.23 % of javascript submissions
  • Your memory usage beats 28.23 % of javascript submissions (35.7 MB)
  • 时间复杂度 O(n^2)

Ⅲ. 遍历法

代码：

/**
 * @param {number} n
 * @return {string}
 */
var countAndSay = function(n) {
    let str = '1'
    for(let i = 1; i < n; i++) {str = countEach(str)
    }
    function countEach(str) {
        let count = 0
        let res = ''
        for(let i = 0; i < str.length; i++) {if(i === 0 || str.charAt(i) === str.charAt(i - 1)) {count++} else {res += `${count}${str.charAt(i - 1)}`
                count = 1
            }
            if (i === str.length - 1) {res += `${count}${str.charAt(i)}`
            }
        }

        return res
    }

    return str
};

结果：

  • 18/18 cases passed (60 ms)
  • Your runtime beats 96.98 % of javascript submissions
  • Your memory usage beats 41.69 % of javascript submissions (35.5 MB)
  • 时间复杂度 O(n^2)

查阅他人解法

这里看到一个开怀一笑的解法，直接字典法，缺点很明显，但是当前情况下确实是最快的。

Ⅰ. 正则法

代码：

/**
 * @param {number} n
 * @return {string}
 */
var countAndSay = function(n) {
    const map = {
        1:"1",
        2:"11",
        3:"21",
        4:"1211",
        5:"111221",
        6:"312211",
        7:"13112221",
        8:"1113213211",
        9:"31131211131221",
        10:"13211311123113112211",
        11:"11131221133112132113212221",
        12:"3113112221232112111312211312113211",
        13:"1321132132111213122112311311222113111221131221",
        14:"11131221131211131231121113112221121321132132211331222113112211",
        15:"311311222113111231131112132112311321322112111312211312111322212311322113212221",
        16:"132113213221133112132113311211131221121321131211132221123113112221131112311332111213211322211312113211",
        17:"11131221131211132221232112111312212321123113112221121113122113111231133221121321132132211331121321231231121113122113322113111221131221",
        18:"31131122211311123113321112131221123113112211121312211213211321322112311311222113311213212322211211131221131211132221232112111312111213111213211231131122212322211331222113112211",
        19:"1321132132211331121321231231121113112221121321132122311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213122112311311123112111331121113122112132113213211121332212311322113212221",
        20:"11131221131211132221232112111312111213111213211231132132211211131221131211221321123113213221123113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331121321231231121113112221121321133112132112312321123113112221121113122113121113123112112322111213211322211312113211",
        21:"311311222113111231133211121312211231131112311211133112111312211213211312111322211231131122211311122122111312211213211312111322211213211321322113311213212322211231131122211311123113223112111311222112132113311213211221121332211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221131112311311121321122112132231121113122113322113111221131221",
        22:"132113213221133112132123123112111311222112132113311213211231232112311311222112111312211311123113322112132113213221133122112231131122211211131221131112311332211211131221131211132221232112111312111213322112132113213221133112132113221321123113213221121113122123211211131221222112112322211231131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122211211133112111311222112111312211312111322211213211321322113311213211331121113122122211211132213211231131122212322211331222113112211",
        23:"111312211312111322212321121113121112131112132112311321322112111312212321121113122112131112131221121321132132211231131122211331121321232221121113122113121113222123112221221321132132211231131122211331121321232221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312211322111312211213211312111322211231131122111213122112311311221132211221121332211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322212321121113122123211231131122113221123113221113122112132113213211121332212311322113212221",
        24:"3113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213213211221113122113121113222112132113213221232112111312111213322112132113213221133112132123123112111311222112132113311213211221121332211231131122211311123113321112131221123113112221132231131122211211131221131112311332211213211321223112111311222112132113212221132221222112112322211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211132221121311121312211213211312111322211213211321322113311213212322211231131122211311123113321112131221123113112211121312211213211321222113222112132113223113112221121113122113121113123112112322111213211322211312113211",
        25:"132113213221133112132123123112111311222112132113311213211231232112311311222112111312211311123113322112132113212231121113112221121321132132211231232112311321322112311311222113111231133221121113122113121113221112131221123113111231121123222112132113213221133112132123123112111312111312212231131122211311123113322112111312211312111322111213122112311311123112112322211211131221131211132221232112111312111213111213211231132132211211131221232112111312212221121123222112132113213221133112132123123112111311222112132113213221132213211321322112311311222113311213212322211211131221131211221321123113213221121113122113121132211332113221122112133221123113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112212211131221121321131211132221123113112221131112311332211211133112111311222112111312211311123113322112111312211312111322212321121113121112133221121321132132211331121321231231121113112221121321132122311211131122211211131221131211322113322112111312211322132113213221123113112221131112311311121321122112132231121113122113322113111221131221",
        26:"1113122113121113222123211211131211121311121321123113213221121113122123211211131221121311121312211213211321322112311311222113311213212322211211131221131211221321123113213221121113122113121113222112131112131221121321131211132221121321132132211331121321232221123113112221131112311322311211131122211213211331121321122112133221121113122113121113222123211211131211121311121321123113111231131122112213211321322113311213212322211231131122211311123113223112111311222112132113311213211221121332211231131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122113221122112133221121113122113121113222123211211131211121311121321123113213221121113122113121113222113221113122113121113222112132113213221232112111312111213322112311311222113111221221113122112132113121113222112311311222113111221132221231221132221222112112322211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322111213122112311311123112112322211213211321322113312211223113112221121113122113111231133221121321132132211331121321232221123123211231132132211231131122211331121321232221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312111213111213211231132132211211131221131211221321123113213221123113112221131112211322212322211231131122211322111312211312111322211213211321322113311213211331121113122122211211132213211231131122212322211331222113112211",
        27:"31131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122211211133112111311222112111312211312111322211213211321322123211211131211121332211231131122211311122122111312211213211312111322211231131122211311123113322112111331121113112221121113122113111231133221121113122113121113222123211211131211121332211213211321322113311213211322132112311321322112111312212321121113122122211211232221123113112221131112311332111213122112311311123112111331121113122112132113311213211321222122111312211312111322212321121113121112133221121321132132211331121321132213211231132132211211131221232112111312212221121123222112132113213221133112132123123112111311222112132113311213211231232112311311222112111312211311123113322112132113212231121113112221121321132122211322212221121123222112311311222113111231133211121312211231131112311211133112111312211213211312111322211231131122211311123113322113223113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331221122311311222112111312211311123113322112132113213221133122211332111213112221133211322112211213322112111312211312111322212321121113121112131112132112311321322112111312212321121113122112131112131221121321132132211231131122211331121321232221121113122113121122132112311321322112111312211312111322211213111213122112132113121113222112132113213221133112132123222112311311222113111231132231121113112221121321133112132112211213322112111312211312111322212311222122132113213221123113112221133112132123222112111312211312111322212321121113121112133221121311121312211213211312111322211213211321322123211211131211121332211213211321322113311213212312311211131122211213211331121321122112133221123113112221131112311332111213122112311311123112111331121113122112132113121113222112311311222113111221221113122112132113121113222112132113213221133122211332111213322112132113213221132231131122211311123113322112111312211312111322212321121113122123211231131122113221123113221113122112132113213211121332212311322113212221",
        28:"13211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322111213122112311311123112112322211213211321322113312211223113112221121113122113111231133221121321132132211331121321232221123123211231132132211231131122211331121321232221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312211322111312211213211312111322211231131122111213122112311311221132211221121332211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221232112111312211312113211223113112221131112311332111213122112311311123112112322211211131221131211132221232112111312211322111312211213211312111322211231131122111213122112311311221132211221121332211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211322113321132211221121332211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321322113311213212322211322132113213221133112132123222112311311222113111231132231121113112221121321133112132112211213322112111312211312111322212311222122132113213221123113112221133112132123222112111312211312111322212311322123123112111321322123122113222122211211232221123113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112212211131221121321131211132221123113112221131112311332211211133112111311222112111312211311123113322112111312211312111322212321121113121112133221121321132132211331121321132213211231132132211211131221232112111312212221121123222112311311222113111231133211121321321122111312211312111322211213211321322123211211131211121332211231131122211311123113321112131221123113111231121123222112111331121113112221121113122113111231133221121113122113121113221112131221123113111231121123222112111312211312111322212321121113121112131112132112311321322112111312212321121113122122211211232221121321132132211331121321231231121113112221121321133112132112312321123113112221121113122113111231133221121321132132211331221122311311222112111312211311123113322112111312211312111322212311322123123112112322211211131221131211132221132213211321322113311213212322211231131122211311123113321112131221123113112211121312211213211321222113222112132113223113112221121113122113121113123112112322111213211322211312113211",
        29:"11131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211132221121311121312211213211312111322211213211321322113311213212322211231131122211311123113223112111311222112132113311213211221121332211211131221131211132221231122212213211321322112311311222113311213212322211211131221131211132221232112111312111213322112131112131221121321131211132221121321132132212321121113121112133221121321132132211331121321231231121113112221121321133112132112211213322112311311222113111231133211121312211231131122211322311311222112111312211311123113322112132113212231121113112221121321132122211322212221121123222112111312211312111322212321121113121112131112132112311321322112111312212321121113122112131112131221121321132132211231131122111213122112311311222113111221131221221321132132211331121321231231121113112221121321133112132112211213322112311311222113111231133211121312211231131122211322311311222112111312211311123113322112132113212231121113112221121321132122211322212221121123222112311311222113111231133211121312211231131112311211133112111312211213211312111322211231131122111213122112311311222112111331121113112221121113122113121113222112132113213221232112111312111213322112311311222113111221221113122112132113121113222112311311222113111221132221231221132221222112112322211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312111322212321121113121112133221132211131221131211132221232112111312111213322112132113213221133112132113221321123113213221121113122123211211131221222112112322211231131122211311123113321112132132112211131221131211132221121321132132212321121113121112133221123113112221131112311332111213211322111213111213211231131211132211121311222113321132211221121332211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322111213122112311311123112112322211213211321322113312211223113112221121113122113111231133221121321132132211331121321232221123123211231132132211231131122211331121321232221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312211322111312211213211312111322211231131122111213122112311311221132211221121332211213211321322113311213212312311211131211131221223113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331121321231231121113112221121321133112132112211213322112312321123113213221123113112221133112132123222112311311222113111231132231121113112221121321133112132112211213322112311311222113111231133211121312211231131112311211133112111312211213211312111322211231131122111213122112311311221132211221121332211211131221131211132221232112111312111213111213211231132132211211131221232112111312211213111213122112132113213221123113112221133112132123222112111312211312111322212311222122132113213221123113112221133112132123222112311311222113111231133211121321132211121311121321122112133221123113112221131112311332211322111312211312111322212321121113121112133221121321132132211331121321231231121113112221121321132122311211131122211211131221131211322113322112111312211322132113213221123113112221131112311311121321122112132231121113122113322113111221131221",
        30:"3113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112212211131221121321131211132221123113112221131112311332211211133112111311222112111312211311123113322112111312211312111322212321121113121112133221121321132132211331121321132213211231132132211211131221232112111312212221121123222112311311222113111231133211121321321122111312211312111322211213211321322123211211131211121332211231131122211311123113321112131221123113111231121123222112111331121113112221121113122113111231133221121113122113121113221112131221123113111231121123222112111312211312111322212321121113121112131112132112311321322112111312212321121113122122211211232221121321132132211331121321231231121113112221121321132132211322132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211322113321132211221121332211231131122211311123113321112131221123113111231121113311211131221121321131211132221123113112211121312211231131122211211133112111311222112111312211312111322211213211321223112111311222112132113213221133122211311221122111312211312111322212321121113121112131112132112311321322112111312212321121113122122211211232221121321132132211331121321231231121113112221121321132132211322132113213221123113112221133112132123222112111312211312112213211231132132211211131221131211322113321132211221121332211213211321322113311213212312311211131122211213211331121321123123211231131122211211131221131112311332211213211321223112111311222112132113213221123123211231132132211231131122211311123113322112111312211312111322111213122112311311123112112322211213211321322113312211223113112221121113122113111231133221121321132132211331222113321112131122211332113221122112133221123113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213122112311311123112112322211322311311222113111231133211121312211231131112311211232221121113122113121113222123211211131221132211131221121321131211132221123113112211121312211231131122113221122112133221121321132132211331121321231231121113121113122122311311222113111231133221121113122113121113221112131221123113111231121123222112132113213221133112132123123112111312211322311211133112111312211213211311123113223112111321322123122113222122211211232221121113122113121113222123211211131211121311121321123113213221121113122123211211131221121311121312211213211321322112311311222113311213212322211211131221131211221321123113213221121113122113121113222112131112131221121321131211132221121321132132211331121321232221123113112221131112311322311211131122211213211331121321122112133221121113122113121113222123112221221321132132211231131122211331121321232221121113122113121113222123211211131211121332211213111213122112132113121113222112132113213221232112111312111213322112132113213221133112132123123112111311222112132113311213211221121332211231131122211311123113321112131221123113112221132231131122211211131221131112311332211213211321223112111311222112132113212221132221222112112322211211131221131211132221232112111312111213111213211231131112311311221122132113213221133112132123222112311311222113111231132231121113112221121321133112132112211213322112111312211312111322212321121113121112131112132112311321322112111312212321121113122122211211232221121311121312211213211312111322211213211321322123211211131211121332211213211321322113311213211322132112311321322112111312212321121113122122211211232221121321132132211331121321231231121113112221121321133112132112312321123113112221121113122113111231133221121321132122311211131122211213211321222113222122211211232221123113112221131112311332111213122112311311123112111331121113122112132113121113222112311311221112131221123113112221121113311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213213211221113122113121113222112132113213221232112111312111213322112132113213221133112132123123112111312211322311211133112111312212221121123222112132113213221133112132123222113223113112221131112311332111213122112311311123112112322211211131221131211132221232112111312111213111213211231132132211211131221131211221321123113213221123113112221131112211322212322211231131122211322111312211312111322211213211321322113311213211331121113122122211211132213211231131122212322211331222113112211"
    }

    return map[n]
};

结果：

  • 18/18 cases passed (56 ms)
  • Your runtime beats 98.81 % of javascript submissions
  • Your memory usage beats 98.5 % of javascript submissions (33.5 MB)
  • 时间复杂度 O(1)

思考总结

总体而言，这道题用正则去解答十分简单，考验的点在正则的匹配这块；当然递归或者遍历也是常规思路；字典法纯属一乐。推荐使用正则来解答此题。

53. 最大子序和

题目地址

题目描述

给定一个整数数组 nums，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。

示例:

输入: [-2,1,-3,4,-1,2,1,-5,4],
输出: 6
解释: 连续子数组 [4,-1,2,1] 的和最大，为 6。

进阶:

如果你已经实现复杂度为 O(n) 的解法，尝试使用更为精妙的分治法求解。

题目分析设想

这道题首先基本解法肯定是暴力的遍历求解，直接遍历找出最大区间。当然这里我们也可以使用动态规划来思考问题，列出动态和转移方程式，等于求解 Max(d[0, i])。另外进阶里面提示分治法，分治法在之前有用过，我们也可以做为一个方向。所以大概有三种：

  • 遍历求解，直接遍历算出各区间值
  • 动态规划问题，求解动态问题，找到每个动态区间的最大值
  • 分治，不断二分找区间内的最大子序和

注意一下，只要是寻找最大值最小值的，初始值需要定义为理论上的最大最小值。

编写代码验证

Ⅰ. 遍历求解

代码：

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxSubArray = function(nums) {
    let res = Number.MIN_SAFE_INTEGER
    for(let i = 0; i < nums.length; i++) {
        let sum = 0
        // 分别算出 i 开始的最大子序和
        for(let j = i; j < nums.length; j++) {sum += nums[j];
            res = Math.max(res, sum)
        }
    }
    return res
};

结果：

  • 202/202 cases passed (272 ms)
  • Your runtime beats 7.61 % of javascript submissions
  • Your memory usage beats 41.88 % of javascript submissions (35.1 MB)
  • 时间复杂度 O(n^2)

Ⅱ. 动态规划

代码：

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxSubArray = function(nums) {let res = dp = nums[0] // 初始值
    for(let i = 1; i < nums.length; i++) {dp = Math.max(dp + nums[i], nums[i]) // 动态取得最大值
        res = Math.max(res, dp)
    }
    return res
};

结果：

  • 202/202 cases passed (68 ms)
  • Your runtime beats 90.14 % of javascript submissions
  • Your memory usage beats 45 % of javascript submissions (35.1 MB)
  • 时间复杂度 O(n)

Ⅲ. 分治

代码：

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxSubArray = function(nums) {
    // 不断分治
    function countByDichotomy (start, end) {
        // 存储左侧结果，右侧结果，两者更大值，以及两者相加的值
        // 解释一下：最大子序列在左右两区间内要么过界要么不过界。// 如果不过界，则最大值为 Max(left, right)
        // 如果过界，则最大为左区间到中间的最大值加中间到右区间的最大值
        if (end === start) { // 数组就一项
            return {lmax: nums[start], // 左半区间包含其右端点的最大子序和
                rmax: nums[start], // 右半区间包含其左端点的最大子序和
                sum: nums[start], // 总和
                result: nums[start] // 区域内部的最大子序和
            }
        } else {const mid = (start + end) >>> 1 // 这个取中位数写法之前解释过，避免溢出
            const left = countByDichotomy(start, mid) // 左区间中计算结果
            const right = countByDichotomy(mid + 1, end) // 右区间中计算结果
            return {lmax: Math.max(left.lmax, left.sum + right.lmax),
                rmax: Math.max(right.rmax, left.rmax + right.sum),
                sum: left.sum + right.sum,
                result: Math.max(left.rmax + right.lmax, Math.max(left.result, right.result))
            }
        }
    }
    return countByDichotomy(0, nums.length - 1).result;
};

结果：

  • 202/202 cases passed (60 ms)
  • Your runtime beats 97.89 % of javascript submissions
  • Your memory usage beats 5.01 % of javascript submissions (36.7 MB)
  • 时间复杂度 O(n)

查阅他人解法

查阅题解的过程中发现了以下几种有意思的思路：

  • 动态规划，使用增益的思路。其实上面我们写的动态规划是一样的
  • 贪心法，尝试多加一位，取较大值
  • 分治中使用贪心法求区间

Ⅰ. 动态规划

代码：

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxSubArray = function(nums) {let res = nums[0]
    let sum = nums[0] // 增益
    for(let i = 1; i < nums.length; i++) {if (sum > 0) { // 正向增益，sum 保留并加上当前遍历数字
            sum += nums[i]
        } else { // sum 更新为当前遍历数字
            sum = nums[i]
        }
        res = Math.max(res, sum)
    }
    return res
};

结果：

  • 202/202 cases passed (60 ms)
  • Your runtime beats 97.89 % of javascript submissions
  • Your memory usage beats 47.5 % of javascript submissions (35.1 MB)
  • 时间复杂度 O(n)

Ⅱ. 贪心法

代码：

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxSubArray = function(nums) {
    let res = Number.MIN_SAFE_INTEGER // 初始值
    let sum = 0
    for(let i = 0; i < nums.length; i++) {sum += nums[i]
        res = Math.max(res, sum)
        if (sum < 0) { // 重新开始找子序串
            sum = 0;
        }
    }
    return res
};

结果：

  • 202/202 cases passed (68 ms)
  • Your runtime beats 90.14 % of javascript submissions
  • Your memory usage beats 45 % of javascript submissions (35.1 MB)
  • 时间复杂度 O(n)

Ⅲ. 分治中使用贪心法求区间

代码：

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxSubArray = function(nums) {
    // 获取跨边界的和
    function getMaxCross (start, mid, end) {
        let leftRes = Number.MIN_SAFE_INTEGER
        let leftSum = 0
        for(let i = mid; i >= start; i--) {leftSum += nums[i]
            leftRes = Math.max(leftRes, leftSum)
        }

        let rightRes = Number.MIN_SAFE_INTEGER
        let rightSum = 0
        for(let i = mid + 1; i <= end; i++) {rightSum += nums[i]
            rightRes = Math.max(rightRes, rightSum)
        }

        return leftRes + rightRes
    }

    function countByDichotomy (start, end) {if (start === end) {return nums[start]
        } else {const mid = (start + end) >>> 1
            const leftSum = countByDichotomy(start, mid)
            const rightSum = countByDichotomy(mid + 1, end)
            const midSum = getMaxCross(start, mid, end)
            // 三者比较最大的就为最大子序和
            return Math.max(leftSum, rightSum, midSum)
        }
    }

    return countByDichotomy(0, nums.length - 1)
};

结果：

  • 202/202 cases passed (72 ms)
  • Your runtime beats 80.56 % of javascript submissions
  • Your memory usage beats 49.38 % of javascript submissions (35.1 MB)
  • 时间复杂度 O(nlog(n))

思考总结

个人认为动态规划在这套题里面解题思路清晰，贪心法也可以理解为基于遍历基础上做的延伸，而分治法需要画图加以理解。一般看到这种最大最长的题目，基本上就可以用动态规划问题来尝试作答了。

58. 最后一个单词的长度

题目地址

题目描述

给定一个仅包含大小写字母和空格 ' ' 的字符串，返回其最后一个单词的长度。

如果不存在最后一个单词，请返回 0。

说明：一个单词是指由字母组成，但不包含任何空格的字符串。

示例：

输入: "Hello World"
输出: 5

题目分析设想

这道题看上去像是一道字符串题，我们可以从以下几个方面来尝试作答：

  • 遍历，从末尾开始，效率高
  • lastIndexOf，直接找空格
  • 正则
  • split

编写代码验证

Ⅰ. 遍历

代码：

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLastWord = function(s) {if(!s.length) return 0
    let i = s.length - 1
    while(i >= 0 && s.charAt(i) === ' ') {i--}
    if(i < 0) return 0 // 全是空格

    let j = i
    while(j >= 0 && s.charAt(j) != ' ') {j--}
    return i - j
};

结果：

  • 59/59 cases passed (64 ms)
  • Your runtime beats 81.14 % of javascript submissions
  • Your memory usage beats 29.72 % of javascript submissions (33.8 MB)
  • 时间复杂度 O(n)

Ⅱ.lastIndexOf

代码：

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLastWord = function(s) {if(!s.length) return 0
    s = s.trim()
    const idx = s.lastIndexOf(' ')
    return idx === -1 ? s.length : s.length - 1 - idx
};

结果：

  • 59/59 cases passed (48 ms)
  • Your runtime beats 99.48 % of javascript submissions
  • Your memory usage beats 36.52 % of javascript submissions (33.7 MB)
  • 时间复杂度 O(1)

Ⅲ. 正则

代码：

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLastWord = function(s) {if(!s.length) return 0
    const match = s.match(/([a-zA-Z]+)\s*$/)
    let res = 0
    if (match) {res = match.pop()
        return res.length
    }
    return res
};

结果：

  • 59/59 cases passed (80 ms)
  • Your runtime beats 26.65 % of javascript submissions
  • Your memory usage beats 5.95 % of javascript submissions (34.2 MB)
  • 时间复杂度 O(1)

Ⅳ.split

代码：

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLastWord = function(s) {if(!s.length) return 0
    s = s.trim()
    const arr = s.split(' ')
    if (arr.length) {let str = arr.pop()
        return str.length
    } else {return 0}
};

结果：

  • 59/59 cases passed (60 ms)
  • Your runtime beats 90.57 % of javascript submissions
  • Your memory usage beats 13.8 % of javascript submissions (34 MB)
  • 时间复杂度 O(1)

查阅他人解法

没有在题解中看到什么特别的解法，大部分都是基于类库解的，比如 Javascript 中 String 和 Array 的方法。或者是遍历实现的。

思考总结

直到现在也没有弄明白这道题的考察点在哪里？不过我建议感兴趣的同学，可以自己拓展实现 lastIndexOf，参考 上一期 28 题的数十种解法，应该能有不小收获。

66. 加一

题目地址

题目描述

给定一个由整数组成的非空数组所表示的非负整数，在该数的基础上加一。

最高位数字存放在数组的首位，数组中每个元素只存储单个数字。

你可以假设除了整数 0 之外，这个整数不会以零开头。

示例：

输入: [1,2,3]
输出: [1,2,4]
解释: 输入数组表示数字 123。输入: [4,3,2,1]
输出: [4,3,2,2]
解释: 输入数组表示数字 4321。

题目分析设想

这道题我有两个大方向，一是数组遍历进行求解，另外一种是数组转数字再处理。但是转数字可能会溢出，所以就只想到从遍历的角度来作答。

  • 遍历，从后往前遍历找到不为 9 的项，后面填 0 就可以了

编写代码验证

Ⅰ. 遍历

代码：

/**
 * @param {number[]} digits
 * @return {number[]}
 */
var plusOne = function(digits) {for(let i = digits.length - 1; i >= 0; i--) {
        // 找不到不为 9 的数，直接加 1 输出就可以了
        if(digits[i] !== 9) {digits[i]++
            return digits
        } else {digits[i] = 0
        }
    }
    digits.unshift(1)
    return digits
};

结果：

  • 109/109 cases passed (60 ms)
  • Your runtime beats 93.72 % of javascript submissions
  • Your memory usage beats 26.35 % of javascript submissions (33.8 MB)
  • 时间复杂度 O(n)

查阅他人解法

发现思路基本都是遍历，但是具体实现会有些差距，这里只列举一个最简单的。

Ⅰ. 遍历

代码：

/**
 * @param {number[]} digits
 * @return {number[]}
 */
var plusOne = function(digits) {for(let i = digits.length - 1; i >= 0; i--) {digits[i]++
        // 取 10 的余数，做了赋值操作，为 0 就继续进位
        digits[i] %= 10
        if(digits[i] !== 0) {return digits}
    }
    digits.unshift(1)
    return digits
};

结果：

  • 109/109 cases passed (64 ms)
  • Your runtime beats 85.29 % of javascript submissions
  • Your memory usage beats 94.34 % of javascript submissions (33.4 MB)
  • 时间复杂度 O(n)

思考总结

这道题可能大众的想法会转成数字再处理，但是做数字运算的时候，千万要记住考虑溢出的问题。另外因为是加 1，所以倒序遍历就可以了。至于是判断末位为 9 还是对 10 取余，我觉得都是一个很好理解的思路，也避免了代码的繁琐。

（完）

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leetcode 算法
发表至： javascript
2019-11-13
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