题目详情
给你一个依照非递加顺序排列的整数数组 nums，和一个目标值 target。请你找出给定目标值在数组中的开始地位和完结地位。
如果数组中不存在目标值 target，返回 [-1, -1]。
你必须设计并实现工夫复杂度为 O(log n) 的算法解决此问题。
示例 1：
** 输出：** nums = [5,7,7,8,8,10], target = 8
** 输入：** [3,4]
示例 2：
** 输出：** nums = [5,7,7,8,8,10], target = 6
** 输入：** [-1,-1]
示例 3：
** 输出：** nums = [], target = 0
** 输入：** [-1,-1]
提醒：
0 <= nums.length <= 105-109 <= nums[i] <= 109nums是一个非递加数组-109 <= target <= 109
解题思路
办法 1 线性查找
首先，咱们从右边对 nums 进行线性扫描，当咱们找到一个指标实例时就会中断。如果咱们从不中断，那么指标就不存在，所以咱们能够提前返回“-1，-1”的“错误代码”。思考到咱们的确找到了一个无效的左索引，咱们能够进行第二次线性扫描，但这次从右开始。在这种状况下，遇到的指标的第一个实例将是最左边的实例（并且因为存在最右边的实例，因而也保障存在最左边的实例）。而后咱们简略地返回一个蕴含两个定位索引的列表。
class Solution {public int[] searchRange(int[] nums, int target) {int[] targetRange = {-1, -1};
// find the index of the leftmost appearance of `target`.
for (int i = 0; i < nums.length; i++) {if (nums[i] == target) {targetRange[0] = i;
break;
}
}
// if the last loop did not find any index, then there is no valid range
// and we return [-1, -1].
if (targetRange[0] == -1) {return targetRange;}
// find the index of the rightmost appearance of `target` (by reverse
// iteration). it is guaranteed to appear.
for (int j = nums.length-1; j >= 0; j--) {if (nums[j] == target) {targetRange[1] = j;
break;
}
}
return targetRange;
}
}办法二 二分查找
除了用于查找左右索引自身的子例程之外，整体算法与线性扫描办法的工作形式十分类似。, 在这里，咱们应用批改后的二分查找来搜寻已排序的数组，并进行一些小的调整。, 首先，因为咱们正在定位蕴含指标的最右边（或最左边）索引（而不是在咱们找到指标时返回 true），所以算法一找到匹配就不会终止。, 相同，咱们持续搜寻直到 lo == hi 并且它们蕴含能够找到指标的索引。
, 另一个变动是引入了左参数，这是一个布尔值，示意在 target == nums [mid] 的事件中要做什么;, 如果右边是真的，那么咱们在关系上的左子数组上“递归”, 否则，咱们在右子数组上递归。要理解为什么这是正确的，请思考咱们在索引 i 处找到指标的状况。, 最右边的指标不能呈现在大于 i 的任何索引处，因而咱们永远不须要思考正确的子阵列。, 雷同的参数实用于最左边的索引。
class Solution {// returns leftmost (or rightmost) index at which `target` should be
// inserted in sorted array `nums` via binary search.
private int extremeInsertionIndex(int[] nums, int target, boolean left) {
int lo = 0;
int hi = nums.length;
while (lo < hi) {int mid = (lo + hi) / 2;
if (nums[mid] > target || (left && target == nums[mid])) {hi = mid;}
else {lo = mid+1;}
}
return lo;
}
public int[] searchRange(int[] nums, int target) {int[] targetRange = {-1, -1};
int leftIdx = extremeInsertionIndex(nums, target, true);
// assert that `leftIdx` is within the array bounds and that `target`
// is actually in `nums`.
if (leftIdx == nums.length || nums[leftIdx] != target) {return targetRange;}
targetRange[0] = leftIdx;
targetRange[1] = extremeInsertionIndex(nums, target, false)-1;
return targetRange;
}
}社区最佳答案
编程语言：java
运行工夫：4ms
战败比例：beat 100%
class Solution {public int[] searchRange(int[] nums, int target) {
int start = 0, end = nums.length - 1;
int[] res = new int[2];
Arrays.fill(res, -1);
while (start <= end) {int mid = start + (end - start) / 2;
if (nums[mid] < target) {start = mid + 1;} else if (nums[mid] > target) {end = mid - 1;}
// if (nums[start] == target && res[0] == -1) {// res[0] = start;
// while (mid <= end) {
//
// }
// }
if (nums[mid] == target) {res[0] = findFirst(nums, start, mid, target);
res[1] = findEnd(nums, mid, end, target);
return res;
// int temp = mid;
// while (start <= mid) {// temp = start + (temp - start) / 2;
// if (nums[temp] < target) {
// start = temp + 1;
// }
//
// if (start == mid
// || (nums[start] == target && (start - 1 < 0 || nums[start - 1] < target))) {// res[0] = start;
// break;
// } else if (start == temp) {
// start++;
// }
// }
//
// temp = mid;
// while (mid <= end) {// temp = temp + (end - temp) / 2;
// if (nums[temp] > target) {
// end = temp - 1;
// }
//
// if (temp == mid
// || nums[temp] == target && (temp + 1 >= nums.length || nums[temp + 1] > target)) {// res[1] = temp;
// break;
// } else if (temp == end) {
// end--;
// }
// }
//
// return res;
}
}
return res;
}
private int findFirst(int[] nums, int start, int end, int target) {while (start < end) {int temp = start + (end - start) / 2;
if (nums[temp] < target) {start = temp + 1;} else if (nums[temp] == target) {end = temp;}
}
return start;
}
private int findEnd(int[] nums, int start, int end, int target) {while (start < end) {int temp = start + (end - start + 1) / 2;
if (nums[temp] > target) {end = temp - 1;} else if (nums[temp] == target) {start = temp;}
}
return start;
}
}编程语言：javascript
运行工夫：52ms
战败比例：beat 100%
/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
let searchRange = function(nums, target) {
let sl = 0;
let sm = 0;
let sr = nums.length - 1;
let el = 0;
let em = 0;
let er = nums.length - 1;
while (sl < sr || el < er) {if (sl < sr) {sm = Math.floor((sl + sr) / 2);
if (nums[sm] < target) {sl = sm + 1;} else {sr = sm;}
}
if (el < er) {em = Math.ceil((el + er) / 2);
if (nums[em] > target) {er = em - 1;} else {el = em;}
}
}
return nums[sl] === target ? [sl, er] : [-1, -1];
};编程语言：python
运行工夫：24ms
战败比例：beat 100%
class Solution(object):
def searchRange(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
if not nums:
return [-1, -1]
left = 0
right = len(nums) - 1
while left <= right:
mid = (left + right) / 2
if nums[mid] == target:
left = mid
right = mid
while left >= 0 and nums[left] == target:
left -= 1
while right <= len(nums) - 1 and nums[right] == target:
right += 1
return [left + 1, right - 1]
elif nums[mid] > target:
right = mid - 1
else:
left = mid + 1
return [-1,-1]
编程语言：python3
运行工夫：40ms
战败比例：beat 100%
class Solution:
def searchRange(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
def binary_find(nums,left,right,k):
res = [-1,-1]
if not nums:return res
while left<right and nums[left]!=nums[right]:
mid = (left+right)//2
if nums[mid]==k:
if nums[left]<k:
left += 1
if nums[right]>k:
right -=1
elif nums[mid]<k:
left = mid+1
else: right = mid-1
if nums[left]==k and nums[right]==k:
return [left,right]
else:return [-1,-1]
return binary_find(nums,0,len(nums)-1,target)
编程语言：cpp
运行工夫：4ms
战败比例：beat 100%
static const auto _____ = []() {ios::sync_with_stdio(false);
cin.tie(nullptr);
return nullptr;
}();
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {vector<int> resultVec(2, -1);
int indexPre;
if ((resultVec[0] = Findpre(nums, target)) != -1) {resultVec[1] = FindLast(nums, target);
}
return resultVec;
}
int Findpre(vector<int>&nums, int target) {
int low = 0;
int high = nums.size() - 1;
while (low <= high) {int middle = (low + high) / 2;
if (nums[middle]<target)
low = middle + 1;
if (nums[middle]>target)
high = middle - 1;
if (nums[middle] == target) {if (middle - 1 == -1 || nums[middle - 1]<target)
return middle;
else {high = middle - 1;}
}
}
return -1;
}
int FindLast(vector<int>&nums, int target) {
int low = 0;
int high = nums.size() - 1;
while (low <= high) {int middle = (low + high) / 2;
if (nums[middle]<target)
low = middle + 1;
if (nums[middle]>target)
high = middle - 1;
if (nums[middle] == target) {if (middle + 1 == nums.size() || nums[middle + 1]>target)
return middle;
else {low = middle + 1;}
}
}
return -1;
}
};编程语言：c
运行工夫：4ms
战败比例：beat 100%
/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int* searchRange(int* nums, int numsSize, int target, int* returnSize) {
int *res;
*returnSize = 2;
res = (int*)malloc(sizeof(int) * 2);
res[0] = -1;
res[1] = -1;
int i;
int flag = 0; // 开始标记
for (i = 0; i < numsSize; i++) {if (nums[i] == target && !flag) {res[0] = i;
flag = 1;
}
if (nums[i] == target && flag &&(i==numsSize-1 || i<numsSize-1 && nums[i+1]>target)) {*(res + 1) = i;
break;
}
}
return res;
}编程语言：swift
运行工夫：16ms
战败比例：beat 100%
class Solution {func searchRange(_ nums: [Int], _ target: Int) -> [Int] {
var left = 0
var right = nums.count - 1
var mid = 0
var first = -1
// 寻找第一个呈现 target 的地位
while left <= right {mid = left + (right - left)/2
if nums[mid] >= target {right = mid - 1} else {left = mid + 1}
if nums[mid] == target {first = mid}
}
// 如果找不到第一个间接返回
if first == -1 {return [first ,first]
}
// 寻找最初一个呈现 target 的地位
var last = -1
left = first
right = nums.count - 1
while left <= right {mid = left + (right - left)/2
if nums[mid] > target {right = mid - 1} else {left = mid + 1}
if nums[mid] == target {last = mid}
}
return [first,last]
}
}参考资料
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