每日一题DFS&每个点都调用一次前后左右由1连接的岛屿数量-211031/220216

Posted 2022-02-16 哥们要飞的blog

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
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

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

本文来自博客园，作者：哥们要飞，转载请注明原文链接：https://www.cnblogs.com/liujinhui/p/15491673.html