【leetcode】Substring with Concatenation of All Words

 1 class Solution {
 2 public:
 3     vector<int> findSubstring(string S, vector<string> &L) {
 4        
 5         if(L.size()==0) return vector<int>();
 6        
 7         int wordLen=L[0].size();
 8        
 9         map<string,int> wordCount;
10        
11         int i,j;
12         for(i=0;i<L.size();i++)  wordCount[L[i]]++;
13        
14        
15         vector<int> result;
16         map<string,int> counting;
17        
18         for(i=0;i<(int)S.length()-wordLen*L.size()+1;i++)
19         {
20             counting.clear();
21             for(j=0;j<L.size();j++)
22             {
23                 string str=S.substr(i+j*wordLen,wordLen);
24                 if(wordCount.find(str)!=wordCount.end())
25                 {
26                     counting[str]++;
27                     if(counting[str]>wordCount[str])
28                     {
29                         break;
30                     }
31                 }
32                 else
33                 {
34                     break;
35                 }
36             }
37            
38             if(j==L.size())
39             {
40                 result.push_back(i);
41                 //i=i+L.size()*wordLen-1;
42             }
43         }
44        
45         return result;
46  
47     }
48 };

 1 class Solution {
 2 public:
 3     vector<int> findSubstring(string S, vector<string> &L) {
 4        
 5         if(L.size()==0) return vector<int>();
 6        
 7         int wordLen=L[0].size();
 8        
 9         map<string,int> wordCount;
10        
11         int i,j;
12         for(i=0;i<L.size();i++)  wordCount[L[i]]++;
13        
14        
15         vector<int> result;
16         map<string,int> counting;
17        
18        
19         int count=0;
20         int start;
21        
22         for(i=0;i<wordLen;i++)
23         {
24             count=0;
25             start=i;
26             counting.clear();
27             for(int j=i;j<S.length()-wordLen+1;j=j+wordLen)
28             {
29                 string str=S.substr(j,wordLen);
30                
31                 if(wordCount.find(str)!=wordCount.end())
32                 {
33                     counting[str]++;
34                     count++;
35                     if(counting[str]>wordCount[str])
36                     {
37                         //更新start位于str第一次出现之后，更新counting，更新count
38                         getNextIndex(start,str,counting,S,wordLen,count);
39                     }
40                 }
41                 else
42                 {
43                     counting.clear();
44                     start=j+wordLen;
45                     count=0;
46                 }
47                
48                 if(count==L.size())
49                 {
50                     result.push_back(start);
51                 }
52             }
53         }
54         return result;
55     }
56    
57    
58     void getNextIndex(int &start,string str,map<string,int> &counting,string &S,int &wordLen,int &count)
59     {
60         for(int j=0;;j++)
61         {
62             string tmpStr=S.substr(start+j*wordLen,wordLen);
63             count--;
64             counting[tmpStr]--;
65             if(tmpStr==str)
66             {
67                 start=start+(j+1)*wordLen;
68                 break;
69             }
70         }
71        
72     }
73 };