200. Number of Islands java solutions-优快云博客

本文介绍了一种基于广度优先搜索(BFS)的算法来计算二维网格中岛屿的数量。算法首先遍历整个网格，当遇到陆地('1')且未被访问过时，通过BFS标记所有相连的陆地，并将岛屿计数加一。该算法适用于由'1'(陆地)和'0'(水域)组成的二维网格。

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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:

Answer: 1

Example 2:

Answer: 3

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

 1 public class Solution {
 2     public int numIslands(char[][] grid) {
 3         if(grid == null || grid.length == 0 || grid[0].length == 0) return 0;
 4         int m = grid.length, n = grid[0].length; 
 5         boolean[][] visit = new boolean[m][n];
 6         int ans = 0;
 7         for(int i = 0; i < m;i++){
 8             for(int j = 0; j < n; j++){
 9                 if(visit[i][j] != true && grid[i][j] == '1'){
10                     ans++;
11                     BFS(visit,grid,i,j);
12                 }
13             }
14         }
15         return ans;
16     }
17     
18     public void BFS(boolean[][] visit, char[][] grid, int row, int col){
19         if(row >=0 && row < visit.length && col >=0 && col < visit[0].length
20             && visit[row][col] != true && grid[row][col] == '1'){
21             visit[row][col] = true;
22             BFS(visit,grid,row-1,col);//上下左右
23             BFS(visit,grid,row+1,col);
24             BFS(visit,grid,row,col-1);
25             BFS(visit,grid,row,col+1);
26         }
27     }
28 }