【38】186. Reverse Words in a String II

186. Reverse Words in a String II

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• Total Accepted: 21466
• Total Submissions: 76192
• Difficulty: Medium
• Contributors: Admin

Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.

The input string does not contain leading or trailing spaces and the words are always separated by a single space.

For example,
Given s = "the sky is blue",
return "blue is sky the".

Could you do it in-place without allocating extra space?

Related problem: Rotate Array

Solution: 

class Solution {
public:
//此题解法：先把每个单词翻转一下，再翻转整个字符串（为了整个单词的大顺序是反过来的）。或者先翻转整个字符串，再翻转每个单词。
    void reverseWords(string &s) {
       reverse(s.begin(), s.end());
        int n = s.length(), l = 0, r = 0;
        while (r < n) {
            while (r < n && !isspace(s[r])) r++;//用isspace判断 可以避免下例的情况hi！
            reverse(s.begin() + l, s.begin() + r); 
            l = ++r;
        }
    }
};

 

以下方法过不了 test case: hi! 输出是!ih 而不是hi！   ？？？？？？？？？？？？？

class Solution {
public:
//此题解法：先把每个单词翻转一下，再翻转整个字符串（为了整个单词的大顺序是反过来的）。或者先翻转整个字符串，再翻转每个单词。
    void reverseWords(string &s) {
        int left = 0;
        for(int i = 0; i <= s.size(); i++){
            if(s[i] == ' ' || i == s.size()){//遇到每个空格或者到最后的时候（因为没空格）要翻转
                //reverse(s, left, i - 1);//从左边界翻转到空格左边，此为一整个单词
                reverse(s.begin() + left, s.begin() + i - 1);
                left = i + 1;//更新左边界，到空格右边第一个
            }
            
        }
        //reverse(s, 0, s.size() - 1);
        reverse(s.begin(), s.end());
    }
    /*
    void reverse(string& s, int left, int right){
        while(left < right){
            char tmp = s[left];
            s[left] = s[right];
            s[right] = tmp;
            left ++; 
            right --;
        }
    }*/
};