200. Number of Islands

Posted 2020-08-29 会咬人的兔子

200. Number of Islands

• Total Accepted: 85982
• Total Submissions: 263924
• Difficulty: Medium
• Contributors: Admin

 

Given a 2d grid map of \'1\'s (land) and \'0\'s (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000

Answer: 1

Example 2:

11000
11000
00100
00011

Answer: 3

DFS:

 1 class Solution {
 2 public:
 3     int numIslands(vector<vector<char>>& grid) {
 4         // Write your code here
 5         if(grid.size() == 0 || grid[0].size() == 0) return 0;
 6         int res = 0;
 7         int m = grid.size();
 8         int n = grid[0].size();
 9         vector<vector<bool>> isVisit(m, vector<bool>(n, false));
10         for(int i = 0; i < m; i++){
11             for(int j = 0; j < n; j++){
12                 if(isVisit[i][j] == false && grid[i][j] == \'1\'){
13                     res++;
14                     dfs(grid, isVisit, i, j);
15                 }
16             }
17         }
18         return res;
19     }
20     void dfs(vector<vector<char>>& grid, vector<vector<bool>>& isVisit, int x, int y){
21         //isVisit[x][y] = true;
22         if (x < 0 || x >= grid.size()) return;
23         if (y < 0 || y >= grid[0].size()) return;
24         if (grid[x][y] != \'1\' || isVisit[x][y]) return;
25         isVisit[x][y] = true;
26         dfs(grid, isVisit, x + 1, y);
27         dfs(grid, isVisit, x - 1, y);
28         dfs(grid, isVisit, x, y + 1);
29         dfs(grid, isVisit, x, y - 1);
30     }
31 };

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