给定一个由 '1'
(陆地)和 '0'
(水)组成的的二维网格,计算岛屿的数量。一个岛被水包围,并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。
示例 1:
输入: 11110 11010 11000 00000 输出: 1
思路:深度优先搜索DFS
public int numIslands(char[][] grid) {
if(grid == null || grid.length == 0 || grid[0].length == 0) return 0;
int m = grid.length;
int n = grid[0].length;
int res = 0;
for(int i = 0; i < m; i++){
for(int j = 0; j < n; i++){
if(grid[i][j] == '1'){
dfsSearch(grid, i, j, m, n);
res++;
}
}
}
return res;
}
public void dfsSearch(char[][] grid, int i, int j, int m, int n){
if(i < 0 || i >= m || j < 0 || j >= n) return;
if(grid[i][j] != '1') return;
grid[i][j] = '0';
dfsSearch(grid, i + 1, j, m, n);
dfsSearch(grid, i - 1, j, m, n);
dfsSearch(grid, i, j + 1, m, n);
dfsSearch(grid, i, j - 1, m, n);
}