Reverse Words in a String【leetcode 1】

leetcode第一题，字符串反转问题：

需要注意的问题：

1.“      a      "只有一个字符是或者字符串最前面有空格时，如何不输出最后那个空格？

2.字符串最后面有空格时如何解决，这个问题在做题目之前发现了，就特意注意了一下

Given an input string, reverse the string word by word.

For example,
Given s = "the sky is blue",
return "blue is sky the".

click to show clarification.

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.

我的思路方法：

读入字符串，并得到长度，每次从最后遍历，若是空格len--；直到遇到非空格，设置一个统计变量c，来统计每个单词的长度，然后在push_back这个单词的没个字符，然后再压入一个空格，如此循环，直到len= 0;

代码带了很多cout<<来调试的

#include <iostream>#include <cstring>#include <cstring>using namespace std;void reverseWords(string &s){int len = s.size();// hello world  11//cout<<"len="<<len<<endl;string ss;while(s[len-1] == ' '){len--;// 去掉最后面的多个空格，如果有空格的话}//cout<<"len2="<<len<<endl;int c = 0;// per word lengthwhile(len > 0){while(s[len-1] == ' ' && len>0){len--;// 去掉（忽略）最后面的多个空格，如果有空格的话}while(s[len - 1] != ' ' && len>0){len--;c++;//单词的长度}//cout<<"len3= "<<len<<"  c="<<c<<endl;for(int i=0;i<c;i++){ss.push_back(s[len+i]); //找到一个单词}while(s[len-1] == ' ' && len>0){len--;// 去掉（忽略）最后面的多个空格，如果有空格的话}if(len>0)ss.push_back(' ');//cout<<"ss="<<ss<<endl;c=0;}s = ss;}int main(){    string s = "   a    ";    cout<<s<<endl;    reverseWords(s);cout<<s<<endl;return 0;}