LeetCode 186. 翻转字符串里的单词 II

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。

class Solution {
public:
    void reverseWords(vector<char>& s) {
        reverse(s, 0, s.size()-1);
        int l = 0, r;
        while(r < s.size())
        {
        	while(r < s.size() && s[r]!=' ')//没遇见分隔符
        		r++;
        	reverse(s, l, r-1);
        	r++;//跳过空格
        	l = r;
        }
    }
    void reverse(vector<char>& s, int l, int r)
    {
    	while(l < r)
    		swap(s[l++],s[r--]);
    }
};