思路

方法：分割 + 倒序

时间复杂度：O(n)，n为s的长度。

 1 class Solution {
 2 public:
 3     string reverseWords(string s) {
 4         string t = "";
 5         stack<string> strStack;
 6         for(int i = 0; i < s.length(); ++i) {
 7             if(s[i] == ' ') {
 8                 if(t != "") {
 9                     strStack.push(t);
10                     t = "";
11                 }
12             } else {
13                 t.push_back(s[i]);
14             }
15         }
16 
17         if(t != "")
18             strStack.push(t);
19 
20         t = "";
21         while(!strStack.empty()) {
22             t += strStack.top();
23             if(strStack.size() > 1) {
24                 t.push_back(' ');
25             } 
26 
27             strStack.pop();
28         }
29 
30         return t;
31     }
32 };