关于leetcode:leetcode之单词替换

序

本文次要记录一下 leetcode 之单词替换

题目

在英语中，咱们有一个叫做 词根 (root) 的概念，它能够跟着其余一些词组成另一个较长的单词——咱们称这个词为 继承词(successor)。例如，词根 an，跟随着单词 other(其余)，能够造成新的单词 another(另一个)。当初，给定一个由许多词根组成的词典和一个句子。你须要将句子中的所有继承词用词根替换掉。如果继承词有许多能够造成它的词根，则用最短的词根替换它。你须要输入替换之后的句子。示例 1：输出：dictionary = ["cat","bat","rat"], sentence = "the cattle was rattled by the battery"
输入："the cat was rat by the bat"

示例 2：输出：dictionary = ["a","b","c"], sentence = "aadsfasf absbs bbab cadsfafs"
输入："a a b c"

示例 3：输出：dictionary = ["a", "aa", "aaa", "aaaa"], sentence = "a aa a aaaa aaa aaa aaa aaaaaa bbb baba ababa"
输入："a a a a a a a a bbb baba a"

示例 4：输出：dictionary = ["catt","cat","bat","rat"], sentence = "the cattle was rattled by the battery"
输入："the cat was rat by the bat"

示例 5：输出：dictionary = ["ac","ab"], sentence = "it is abnormal that this solution is accepted"
输入："it is ab that this solution is ac"

 

提醒：1 <= dictionary.length <= 1000
    1 <= dictionary[i].length <= 100
    dictionary[i] 仅由小写字母组成。1 <= sentence.length <= 10^6
    sentence 仅由小写字母和空格组成。sentence 中单词的总量在范畴 [1, 1000] 内。sentence 中每个单词的长度在范畴 [1, 1000] 内。sentence 中单词之间由一个空格隔开。sentence 没有前导或尾随空格。起源：力扣（LeetCode）链接：https://leetcode-cn.com/problems/replace-words
著作权归领扣网络所有。商业转载请分割官网受权，非商业转载请注明出处。

题解

class Solution {public String replaceWords(List<String> dictionary, String sentence) {String[] words = sentence.split("\\s+");
        for (int i = 0; i < words.length; i++) {for (String dic : dictionary) {if (words[i].startsWith(dic)) {words[i] = dic;
                }
            }
        }
        return Arrays.asList(words)
                    .stream()
                    .collect(Collectors.joining(" "));
    }          
}

小结

这里用双层循环应用 startsWith 来判断是否命中词根，如果是则替换，如果后面命中的词根不是最短的，则前面遇到会被替换掉，最初再将替换后的 words 数组拼接为 sentence。

doc

• 单词替换