问题的关键与如何判断一个岛屿。

岛屿是连通的,那么通过深度遍历或者宽度遍历将遍历的过的节点进行标记,当不能标记的时候说明这个岛屿遍历结束。

import java.util.*;

public class Solution {
    /**
     * 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
        int count = 0;
        
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                if (grid[i][j] == '0') {
                    continue;
                } else {
                    count++;
                    deepSearch(grid, i, j);
                }
            }
        }
        return count;
    }
    
    private void deepSearch(char[][] grid, int i, int j) {
        if (i >= grid.length || i < 0) {
            return;
        }
        if (j >= grid[i].length || j < 0) {
            return;
        }
        if (grid[i][j] == '0') {
            return;
        }
        if (grid[i][j] == '1') {
            grid[i][j] = '0';
        }
        deepSearch(grid, i + 1, j);
        deepSearch(grid, i - 1, j);
        deepSearch(grid, i, j + 1);
        deepSearch(grid, i, j - 1);
    }
}