LeetCode--Substring with Concatenation of All Words

通过hash记录L中各个单词出现的次数，由于每个单词仅仅在S的解中出现一次，且单词长度一样，所以还是很好处理的。

但是有一个问题需要注意：

string的length函数返回的size_t是无符号的。参看代码注释1

http://www.cplusplus.com/reference/string/string/length/

 1 class Solution {
 2 public:
 3     vector<int> findSubstring(string S, vector<string> &L) {
 4         int lSize = L.size();
 5         vector<int> res;
 6         if(lSize == 0)
 7         {
 8             return res;
 9         }
10 
11         map<string,int> countL;
12         int i;
13         for(i = 0 ; i < lSize ; ++i)
14         {
15             if(countL.find(L[i]) != countL.end())
16                 countL[L[i]]++;
17             else
18                 countL[L[i]] = 1;
19         }
20         map<string,int> countS;
21         int wordSize = L[0].length();
22         //cout<<S.length()<<lSize<<wordSize<<endl;
23         int len = S.length();
24         //cout<<len-(lSize*wordSize)<<endl;
25         for(i = 0 ; i <= len-lSize*wordSize;++i)
//注释1：这里一开始我是直接写的S.length()替代len的，但是当S.length()小于lsize*wordsize的时候，就会出现返回的无符号整型很大很大
26         {
27             int j;
28             countS.clear();
29             for(j = 0 ; j < lSize ; ++j)
30             {
31                 string word = S.substr(i+j*wordSize,wordSize);
32                 if(countS.find(word) != countS.end())
33                 {
34                     ++countS[word];
35                     if(countS[word] > countL[word])
36                         break;
37                 }
38                 else if(countL.find(word) != countL.end())
39                 {
40                     countS[word] = 1;
41                 }
42                 else
43                     break;
44             }
45             if(j == lSize)
46             {
47                 res.push_back(i);
48             }
49         }
50         return res;
51     }
52 };