Description:
Given an input string, reverse the string word by word.
For example,
Given s = "the sky is blue",
return "blue is sky the".
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.
分析： 根据题目中的解释，基本就可以弄清这个题目的基本要求. 翻转字符串中的单词。 基本想法是用stack存所有的单词，然后重建字符串。
代码中的注释有分析，后面那个runtime error告诉我们几种极端测试检查是需要的！
然后注意那个常用trick，在处理一些边界条件时还是很有用的
1 class Solution { 2 public: 3 void reverseWords(string &s) { 4 stack<string> rec; 5 if(s.empty()) return; 6 //这里有这个trick,在最后加一个标志字符，象征结束，免得还要在后面循环结束后，看是否还有 7 //单词没有加入stack中,常用trick 8 s = s+" "; 9 string onec; 10 int startind = -1; 11 for(int i=0;i<s.size();i++) 12 { 13 if(s[i]==' ' && startind==-1) continue; 14 else if(s[i]==' '){ 15 onec.assign(s.begin()+startind,s.begin()+i); 16 rec.push(onec); 17 onec.clear(); 18 startind = -1; 19 continue; 20 } 21 else if (startind==-1) startind=i; 22 23 } 24 s=""; 25 //必须判断rec是否为空，否则会runtime error，因为后面的删除操作 26 if(rec.empty()) return; 27 while(!rec.empty()) 28 { 29 s = s+ rec.top()+" "; 30 rec.pop(); 31 } 32 s.erase(s.end()-1); 33 34 } 35 };