LeetCode题解——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".

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.

解题思路：如果不需要inplace，那么可以借助于stringstream和vector或者substr等操作，可以很容易的实现reverse，但是如果要用inplace的翻转的话，需要新的想法。

分为两步：第一步翻转句子中的所有字符串，第二步翻转每个单词的顺序。在此过程中要考虑处理连续的空字符。

AC代码如下：

class Solution {public:    void reverseWords2(string &s,string::iterator begin,string::iterator end){         if(!*begin || !*end) return;         while(begin<end){char temp = *begin;            *begin = *end;* end = temp;++begin;--end;        }    }void reverseWords(string &s) {while(s.size() && *s.begin()==' ') s.erase(s.begin());if(!s.size()) return;        reverseWords2(s,s.begin(),s.end()-1);string::iterator  begin = s.begin(), end =s.begin();while(begin!=s.end()-1){if(*begin==' '){ s.erase(begin);}//++begin; ++end;}else if(end == s.end() || *end ==' '){reverseWords2(s,begin,end-1);if(end == s.end()) return;begin = ++end;}else{++end;}}    }    //erase操作耗时};