Reverse Words in a String -- leetcode

Given an input string, reverse the string word by word.

For example,

Given s = "

the sky is blue

",

return "

blue is sky the

".

Update (2015-02-12):

For C programmers: Try to solve it in-place in O(1) space.

click to show clarification.

Clarification:

What constitutes a word?

A sequence of non-space characters constitutes a word.
Could the input string contain leading or trailing spaces?

Yes. However, your reversed string should not contain leading or trailing spaces.
How about multiple spaces between two words?

Reduce them to a single space in the reversed string.

基本思路：

题目要求，将句子出现的单词顺序逆转一下，但单词本身不逆转。

1. 先将字符串整体reverse；

2. 再将每个单词作一下reverse。

3. 由于存在前导空格，单词之间，多个空格的情况。 需要将单词向前作移动，以压缩空格。

以下代码中，变量p表示，处理已经完成的末尾。也即下一个单词移动的目标位置。

变量q表示，下一个单词的开始位置。

此代码在leetcode上实际执行时间为8ms。

class Solution {
public:
    void reverseWords(string &s) {
        reverse(s.begin(), s.end());
        int p = 0;
        int q = 0;
        for (int i=0; i<s.size(); i++) {
            if (s[i] == ' ') {
                if (i == q)
                    ++q;
                else {
                    reverse(s.begin()+q, s.begin()+i);
                    while (q <= i) s[p++] = s[q++];
                }
            }
        }
        reverse(s.begin()+q, s.end());
        while (q < s.size()) s[p++] = s[q++];
        s.erase(p);
        if (!s.empty() && s[s.size()-1] == ' ')
            s.pop_back();
    }
};