【leetcode】Word Search

 1 class Solution {
 2 public:
 3  
 4  
 5     int n,n1,n2;
 6  
 7    
 8     bool exist(vector<vector<char> > &board, string word) {
 9        
10         n1=board.size();
11         n2=board[0].size();
12         n=word.length();
13        
14         bool flag=false;
15        
16         //vector<vector<bool> > visited(n1,vector<bool>(n2,false));
17        
18         bool **visited=new bool*[n1];
19         for(int i=0;i<n1;i++)
20         {
21             visited[i]=new bool[n2];
22             for(int j=0;j<n2;j++)
23             {
24                 visited[i][j]=false;
25             }
26         }
27        
28        
29        
30         for(int i=0;i<n1;i++)
31         {
32             for(int j=0;j<n2;j++)
33             {
34                 if(board[i][j]==word[0])
35                 {
36                     //注意visited采用引用传值
37                     flag=flag||dfs(board,word,i,j,visited);
38                     if(flag) return true;
39                 }
40             }
41         }
42        
43        
44         for(int i=0;i<n1;i++) delete[] visited[i];
45        
46         return false;
47     }
48    
49  
50     bool dfs(vector<vector<char> > &board,string &word,int i,int j,bool** &visited,int index=0)
51     {
52  
53         if(board[i][j]!=word[index]||visited[i][j]) return false;
54         if(index==n-1) return true;
55        
56         visited[i][j]=true;
57        
58         bool flag1=i+1<n1&&dfs(board,word,i+1,j,visited,index+1);
59         bool flag2=j+1<n2&&dfs(board,word,i,j+1,visited,index+1);
60         bool flag3=j-1>=0&&dfs(board,word,i,j-1,visited,index+1);
61         bool flag4=i-1>=0&&dfs(board,word,i-1,j,visited,index+1);
62        
63         bool result=flag1||flag2||flag3||flag4;
64        
65         //由于是引用传值，所以没有找到的目标串时要把visited复原
66         //if(result==false) visited[i][j]=false;         
67         visited[i][j]=result;
68         return result;          
69     }
70 };