翻转字符串里的单词 II

问题描述

给定一个字符串，逐个翻转字符串中的每个单词。

示例：
输入: ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
输出: ["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]

注意：
单词的定义是不包含空格的一系列字符
输入字符串中不会包含前置或尾随的空格
单词与单词之间永远是以单个空格隔开的
进阶：使用 O(1) 额外空间复杂度的原地解法。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-words-in-a-string-ii

思路

• 以空格为分隔，将每个单词提取出来。
• 按照单词之间的倒序顺序，但是单词内部的正常顺序，将单词存到ret中。
• 最后，返回ret即可。

题解

class Solution {
public:
    void reverseWords(vector<char> &s) {
        vector<char> ret;
        string tmp;
        vector<string> tmp1;
        for (int i = 0; i < s.size(); i++) {
            if (s[i] != ' ') {
                tmp += s[i];
            } else if (s[i] == ' ') {
                tmp1.push_back(tmp);
                tmp = "";
            }
        }
        tmp1.push_back(tmp);
        reverse(tmp1.begin(), tmp1.end());
        for (int i = 0; i < tmp1.size(); i++) {
            for (int j = 0; j < tmp1[i].length(); j++) {
                ret.push_back(tmp1[i][j]);
            }
            if (i != tmp1.size() - 1) {
                ret.push_back(' ');
            }
        }
        s=ret;
    }
};