题目

• 原题连接：200. 岛屿数量

1-思路

• 利用 dfs 深搜，遇到岛屿直接将岛屿填充为 0

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;
        }

        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实现

public class islandDFS {

    public static 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 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 scanner = new Scanner(System.in);
        int m = scanner.nextInt(); // 读取行数
        int n = scanner.nextInt(); // 读取列数
        scanner.nextLine(); // 跳过行末的换行符

        char[][] grid = new char[m][n];
        for (int i = 0; i < m; i++) {
            String line = scanner.nextLine();
            String[] numbers = line.split(" ");
            for (int j = 0; j < n; j++) {
                grid[i][j] = numbers[j].charAt(0);
            }
        }

        int count = numIslands(grid);
        System.out.println(count);

    }
}s