Character shortest path solution for leetCode problem 2

1. Description of the topic

2. Analysis

　　The algorithm used before to find both sides of the big loop is O(n^2) complexity, and the complexity can be reduced by using multimap.

3. Code

1 vector< int > shortestToChar( string S, char C) {
 2          // use the multimap in the standard library to store each character and its subscript
 3          // the advantage of multimap is that the key value can be repeated 
4          vector< int > ans;
 5          multimap< char , int > m;
 6          for ( int i = 0 ; i< S.size(); i++ )
 7              m.insert( make_pair(S[i],i) );
 8          
9          for ( int i = 0 ;i<S.size();i++)
10         {
11             if( S[i] == C )
12             {
13                 ans.push_back(0);
14             }    
15             else
16             {
17                 int k = S.size()-1;
18                 for( auto pos = m.lower_bound( C ); pos != m.upper_bound( C ); pos++ )
19                 {
20                     
21                         if( abs( pos->second - i ) < k )
22                             k = abs( pos->second -i );
23                 }
24                  ans.push_back(k); 
25             } 
26         }
27         return ans; 
28     }