---

• 题目连接：200. 岛屿数量

---

1- 思路

DFS 深搜

• 在遍历中对 res 结果进行 ++ 操作 。
• 遇到一个陆地结果为 1 的地方， 就将他们直接填充为 0

思路

• ① 先遍历，收集 res
• ② 之后通过 dfs 遇到岛屿，直接填充为 '0'

深搜三部曲

• 递归参数返回值
• 确认终止条件

---

2- 实现

⭐200. 岛屿数量——题解思路

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'){res++;dfs(grid,i,j);}}}return res;}public void dfs(char[][] grid,int i ,int j){// 终止条件if(i<0 || i >= grid.length || j<0 || j>= grid[0].length || grid[i][j] == '0'){return ;}// 填充grid[i][j] = '0';dfs(grid,i-1,j);dfs(grid,i+1,j);dfs(grid,i,j+1);dfs(grid,i,j-1);}
}

---

3- ACM 思路

public class numIslands {public static int numIsland(char[][] grid){int res = 0;int m = grid.length;int n = grid[0].length;for(int i = 0 ; i < m ;i++){for(int j = 0 ; j < n;j++){if(grid[i][j] == '1'){res++;dfs(grid,i,j);}}}return res;}public static void dfs(char[][] grid,int i ,int j){// 终止条件if(i<0 || i>= grid.length || j<0 || j>grid[0].length || grid[i][j] =='0'){return ;}grid[i][j] = '0';dfs(grid,i-1,j);dfs(grid,i+1,j);dfs(grid,i,j-1);dfs(grid,i,j+1);}public static void main(String[] args) {Scanner sc = new Scanner(System.in);String input = sc.nextLine();input = input.substring(2, input.length()-2).replace("\"","");String[] rows = input.split("\\],\\[");int m = rows.length;int n = rows[0].split(",").length;char[][] grid = new char[m][n];int rIndex = 0;for(String str : rows){String[] row = str.split(",");for(int j = 0 ;  j < n;j++){grid[rIndex][j] = row[j].charAt(0);}rIndex++;}System.out.println("结果是"+numIsland(grid));}}