[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.

这道题要求将一个字符串中的单词顺序反转，单词内部不反转，同时将单词间的多个空格合并成一个，头尾的空格删除。

思路：

1. 将整个字符串按照字母整个进行反转

2. 单词内进行反转，设置first和last分别表示单词的结束和开始，N个空格的第一个的前一个位置是单词的结尾，N个空格的最后一个的下一个是单词的开始。同时删去多余的空格。

3. 不额外使用存储空间。

下面贴上代码：

class Solution {
public:
    void reverseWords(string &s) {
        reverse(s.begin(), s.end());
        int first=0;
        int last=0;
        int i = 0;
        while (first != s.length()){
            while (first != s.length() && s[first] == ' ')
                first++;
            last = first;
            while (last != s.length() && s[last] != ' ')
                last++;
            if (i != 0 && first <= last - 1)
                s[i++] = ' ';
            for (int j = last - 1; first < j; first++, j--){
                swap(s[j], s[first]);
                s[i++] = s[first];
            }
            while (first < last){
                s[i++] = s[first++];
            }
        }
        s.resize(i);
    }
};