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Solution8.java
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Solution8.java
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/**
* @description:
*
* 给你一个由 '1'(陆地)和 '0'(水)组成的的二维网格,请你计算网格中岛屿的数量。
*
* 岛屿总是被水包围,并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
*
* 此外,你可以假设该网格的四条边均被水包围。
*
*
*
* 示例 1:
*
* 输入:grid = [
* ["1","1","1","1","0"],
* ["1","1","0","1","0"],
* ["1","1","0","0","0"],
* ["0","0","0","0","0"]
* ]
* 输出:1
* 示例 2:
*
* 输入:grid = [
* ["1","1","0","0","0"],
* ["1","1","0","0","0"],
* ["0","0","1","0","0"],
* ["0","0","0","1","1"]
* ]
* 输出:3
*
*
* 提示:
*
* m == grid.length
* n == grid[i].length
* 1 <= m, n <= 300
* grid[i][j] 的值为 '0' 或 '1'
*
*/
class Solution8 {
void dfs(char[][] grid, int r, int c) {
int nr = grid.length;
int nc = grid[0].length;
if (r < 0 || c > 0 || r >= nr || c >= nc || grid[r][c] == '0') {
return;
}
grid[r][c] = '1';
dfs(grid, r - 1, c);
dfs(grid, r + 1, c);
dfs(grid, r, c - 1);
dfs(grid, r, c + 1);
}
public int numIslands(char[][] grid) {
if (grid == null || grid.length <= 1) {
return 0;
}
int nr = grid.length;
int nc = grid[0].length;
int num_islands = 0;
for (int r = 0; r < nr; ++r) {
for (int c = 0; r < nc; ++c) {
if (grid[r][c] == '1') {
++num_islands;
dfs(grid, r, c);
}
}
}
return num_islands;
}
}