[LeetCode] 200. Number of Islands 解题思路

 1     unordered_map<string, pair<int, int>> val_pos;
 2     vector<vector<char>> theGrid;
 3 
 4 
 5     /**
 6      * remove all land connecting with [i, k] adjecent horizontally or vertically. and the pos.
 7      */
 8     void removeLand(int i, int k){
 9         
10         theGrid[i][k] = ‘0‘;
11         
12         string str = to_string(i) + "_" + to_string(k);
13         val_pos.erase(str);
14         
15         if (i-1 >= 0 && theGrid[i-1][k] == ‘1‘){
16             removeLand(i-1, k);
17         }
18         
19         if (k+1 < theGrid[0].size() && theGrid[i][k+1] == ‘1‘){
20             removeLand(i, k+1);
21         }
22         
23         if (i+1 < theGrid.size() && theGrid[i+1][k] == ‘1‘){
24             removeLand(i+1, k);
25         }
26         
27         if (k-1 >= 0 && theGrid[i][k-1] == ‘1‘){
28             removeLand(i, k-1);
29         }
30     }
31 
32     int numIslands(vector<vector<char>>& grid) {
33         
34         if (grid.size() == 0){
35             return 0;
36         }
37 
38         theGrid = grid;
39         for ( int i = 0 ; i < grid.size(); i++){
40             for(int k = 0 ; k < grid[0].size(); k++){
41                 if (grid[i][k] == ‘1‘){
42                     string str = to_string(i) + "_" + to_string(k);
43                     val_pos[str] = {i, k};
44                 }
45             }
46         }
47         
48         int res = 0;
49         
50         while(val_pos.size() > 0 ){
51             int i = val_pos.begin()->second.first;
52             int k = val_pos.begin()->second.second;
53             removeLand(i, k);
54             res++;
55         }
56         
57         return res;
58     }