[LeetCode] 186. Reverse Words in a String II 翻转字符串中的单词 II

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?

151. Reverse Words in a String 一样，这里要求不能用额外空间，要用in-place完成。

该题假设开头和结尾没有空格，而且单词之间只有一个空格。不需要这些假设也是可以的，就是代码会比较复杂。
思路就是两步走，第一步就是将整个字符串翻转。然后从头逐步扫描，将每个遇到单词再翻转过来。

[注意事项]
1）如果是Java，应该跟面试官指出String是immutable，所以需要用char array来做。
2）follow-up问题：k-step reverse。也就是在第二部翻转的时候，把k个单词看作一个长单词，进行翻转。

Java：

public void reverseWords(char[] s) {  
    reverse(s, 0, s.length);  
    for (int i=0, j=0; j<=s.length; j++) {  
        if (j==s.length || s[j]==' ') {  
            reverse(s, i, j);  
            i =  j + 1;  
        }  
    }  
}  
   
private void reverse(char [] s, int begin, int end) {  
    for (int i=0; i<(end-begin)/2; i++) {  
        char temp = s[begin+i];  
        s[begin+i] = s[end-i-1];  
        s[end-i-1] = temp;  
    }  
} 

Python:

class Solution(object):
    def reverseWords(self, s):
 
        def reverse(s, begin, end):
            for i in xrange((end - begin) / 2):
                s[begin + i], s[end - 1 - i] = s[end - 1 - i], s[begin + i]
 
        reverse(s, 0, len(s))
        i = 0
        for j in xrange(len(s) + 1):
            if j == len(s) or s[j] == ' ':
                reverse(s, i, j)
                i = j + 1

C++:

class Solution {
public:
    void reverseWords(string &s) {
        int left = 0;
        for (int i = 0; i <= s.size(); ++i) {
            if (i == s.size() || s[i] == ' ') {
                reverse(s, left, i - 1);
                left = i + 1;
            }
        }
        reverse(s, 0, s.size() - 1);
    }
    void reverse(string &s, int left, int right) {
        while (left < right) {
            char t = s[left];
            s[left] = s[right];
            s[right] = t;
            ++left; --right;
        }
    }
};　　