substring with concatenation 

 1 class Solution {
 2 public:
 3     vector<int> findSubstring(string S, vector<string> &L) {
 4         // Start typing your C/C++ solution below
 5         // DO NOT write int main() function
 6         vector<int> v;
 7         if( L.empty() ) return v;
 8         int wn = L.size();
 9         int wl = L[0].size();
10         if( S.length() < wn*wl ) return v;
11         map<string,int> count;
12         for(int i=0;i<wn;i++)
13         {
14             if( count.find(L[i])!=count.end())
15                 count[L[i]] ++;
16             else
17                 count[L[i]] = 1;
18         }
19         map<string,int> found;
20         int fn;
21         for(int i=0;i<wl;i++)
22         {
23             fn = 0;
24             for(int j=i;j+wl<=S.length();j+=wl)
25             {
26                 if( j+wl*(wn-fn) > S.length() )
27                     break;
28                 string str = S.substr( j,wl);
29                 if( count.find(str)!=count.end() )
30                 {
31                     if(found.find(str)!=found.end())
32                     {
33                         found[str] ++;
34                         if( found[str] > count[str] )
35                         {
36                             int k = j - wl*fn;
37                             string s;
38                             while(k<j && (s = S.substr( k,wl))!=str)
39                             {
40                                 found[s] --;
41                                 fn --;
42                                 if( found[s] == 0 )
43                                     found.erase( s );
44                                 k += wl;
45                             }
46                             found[str] --;                          
47                         }   
48                         else
49                             fn++;
50                         if( fn == wn )
51                             v.push_back( j - (wn-1)*wl );                       
52                     }
53 
54                     else
55                     {
56                         found[str] = 1;
57                         fn++;
58                         if( fn == wn )
59                             v.push_back( j - (wn-1)*wl );
60                     }
61                 }
62                 else
63                 {
64                     found.clear();
65                     fn = 0;
66                  }
67 
68             }
69         }
70         return v;
71 
72     }
73 };

 

posted on 2013-09-06 17:05  jumping_grass  阅读(162)  评论(0)    收藏  举报

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