题解 | #岛屿数量#

import java.util.*;


public class Solution {

    /**
    m,n表示坐标,把与m和n相邻的1 都置为 0
    */
    static void dfs(char[][] grid, int m, int n) {
        if (m < 0 || n < 0 || m > grid.length || n > grid[0].length) {
            return;
        }
        grid[m][n] = '0';
        if (m > 0 && m < grid.length && grid[m - 1][n] == '1') {
            dfs(grid, m - 1, n);
        }
        if (m >= 0 && m < grid.length - 1 && grid[m + 1][n] == '1') {
            dfs(grid, m + 1, n);
        }
        if (n > 0 && n < grid[0].length && grid[m][n - 1] == '1') {
            dfs(grid, m, n - 1);
        }
        if (n >= 0 && n < grid[0].length - 1 && grid[m][n + 1] == '1') {
            dfs(grid, m, n + 1);
        }
    }

    /**
     * 判断岛屿数量
     * @param grid char字符型二维数组
     * @return int整型
     */
    public int solve (char[][] grid) {
        // write code here
        //空矩阵
        if (grid.length == 0) {
            return 0;
        }
        int count = 0;//岛屿的数量
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == '1') {
                    count++;
                    dfs(grid, i, j);
                }
            }
        }
        return count;
    }
}