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
};



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