[LeetCode] Reverse Words in a String

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.

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.

反转字符串中单词的顺序。有3中可能的情况（用_表示空格）：

1. a_b

2. __

3. _ab

4. ab_

第一种情况也是最普遍的情况，检测空格并反转字符串中单词即可。

第二、三、四种情况需要消除由1反转后多余空格。思路是将二、三种情况都转化为第四种，利用一个变量计数来统计非空格的字符数量，使用substr将最后一个空格消去

class Solution {
public:
    void reverseWords(string &s) {
        reverse(s.begin(), s.end());
        int tmp = 0;
        for (int i = 0; i != s.size(); i++) {
            if (s[i] == ' ') {
                reverse(s.begin() + tmp, s.begin() + i);
                tmp = i + 1;
            }
        }
        reverse(s.begin() + tmp, s.end());
        int t = 0;
        for (int i = 0; i != s.size(); i++) {
            if (s[i] != ' ') {
                s[t++] = s[i];
                if (s[i + 1] == ' ' || i == s.size() - 1)
                    s[t++] = ' ';
            }
        }
        s = s.substr(0, t == 0 ? 0 : t - 1);
    }
};
// 6 ms