[leetcode] 151. 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.

这道题是倒序输出字符串中单词，题目难度为Medium。

最直观的想法是将单词逐个存入栈中，然后依次出栈生成倒序后的字符串。具体代码：

class Solution {public:    void reverseWords(string &s) {        stack<string> words;        string curWord = "";                s += " ";        for(auto ch:s) {            if(ch == ' ') {                if(!curWord.empty()) {                    words.push(curWord);                    curWord = "";                }            }            else curWord += ch;        }                s = "";        while(!words.empty()) {            s += words.top();            words.pop();            if(!words.empty()) s += " ";        }    }};

不过题目要求in-place，所以栈不能用了。我们可以先逐个倒序每个单词，然后再将整个字符串倒序，这样也能够得到所求结果，同时保证了in-place。具体代码：

class Solution {public:    void reverseWords(string &s) {        int idx = 0, pos = 0, len = 0;                while(true) {            while(idx < s.size() && s[idx] == ' ') ++idx;            if(idx == s.size()) break;            if(pos) s[pos++] = ' ';            while(idx < s.size() && s[idx] != ' ') s[pos+(len++)] = s[idx++];            reverse(s.begin()+pos, s.begin()+pos+len);            pos += len;            len = 0;        }                s = s.substr(0, pos);        reverse(s.begin(), s.end());    }};