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
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/number-of-islands
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思路：深度优先搜索DFS
方法1：使用visited数组记录是否被访问过☆
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
*/
//DFS
public int solve (char[][] grid) {
int res = 0;
int m = grid.length, n = grid[0].length;
boolean[][] visited = new boolean[m][n];
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == '1' && visited[i][j] == false) {
dfs(grid, i, j, visited);
res++;
}
}
}
return res;
}
public void dfs(char[][] grid, int i, int j, boolean[][] visited) {
if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || visited[i][j] || grid[i][j] == '0') {
return;
}
visited[i][j] = true;
dfs(grid, i - 1, j, visited);
dfs(grid, i, j - 1, visited);
dfs(grid, i + 1, j, visited);
dfs(grid, i, j + 1, visited);
}
}
方法2：将访问过的标为2
深度优先搜索
class Solution {
public int numIslands(char[][] grid) {
int res = 0;
for(int i = 0; i < grid.length; i++) {
for(int j = 0; j < grid[0].length; j ++) {
if(grid[i][j] == '1') {
dfs(grid, i, j);
res ++;
}
}
}
return res;
} public void dfs(char[][] grid, int i, int j) {
if(i < 0 || i > grid.length - 1 || j < 0 || j > grid[0].length - 1 || grid[i][j] != '1') {
return;
}
grid[i][j] = '2';
dfs(grid, i - 1, j);
dfs(grid, i + 1, j);
dfs(grid, i, j - 1);
dfs(grid, i, j + 1);
}
}
类似题目：N皇后等
