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。

思路：先反转每个单词，然后总体再翻转。

class Solution {
    public void reverseWords(char[] s) {
        int start=0;
        for(int i=0;i<s.length;i++){
            if(s[i]==' '){
                reverseWord(s,start,i-1);
                start=i+1;
            }
        }
        reverseWord(s,start,s.length-1);
        reverseWord(s,0,s.length-1);
    }
    public void reverseWord(char[] s,int start,int end){
        char temp;
        while(start<end){
            temp=s[start];
            s[start]=s[end];
            s[end]=temp;
            start++;
            end--;
        }
    }
}