【Java】力扣_每日一题_200. 岛屿数量_中等

题目链接:https://leetcode-cn.com/problems/number-of-islands/

题目描述
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。

第一次编辑代码:
(他写垃圾算法一直可以的)

class Solution {
    public int numIslands(char[][] grid) {
        if(grid == null || grid.length == 0)
            return 0;
            
        int row = grid.length, col = grid[0].length;
        int ans = 0;
        for(int i = 0; i < row; i++)
            for(int j = 0; j < col; j++){
                if(grid[i][j] == '1'){   
                    dfs(i, j, grid, row, col);
                    ans++;
                }
            }
        return ans;
    }

    public void dfs(int i, int j, char[][] grid, int row, int col){
        
        grid[i][j] = '2';

        if(i-1 >= 0 && grid[i-1][j] == '1')
            dfs(i-1, j, grid, row, col);
        if(i+1 < row && grid[i+1][j] == '1')
            dfs(i+1, j, grid, row, col);
        if(j-1 >= 0 && grid[i][j-1] == '1')
            dfs(i, j-1, grid, row, col);
        if(j+1 < col && grid[i][j+1] == '1')
            dfs(i, j+1, grid, row, col);
    }
}

反思
太蠢了,判断条件可以提出来不用重复写。

第二次编辑代码:

class Solution {
    public int numIslands(char[][] grid) {
        if(grid == null || grid.length == 0)
            return 0;

        int row = grid.length, col = grid[0].length;
        int ans = 0;
        for(int i = 0; i < row; i++)
            for(int j = 0; j < col; j++){
                if(grid[i][j] == '1'){   
                    dfs(i, j, grid, row, col);
                    ans++;
                }
            }
        return ans;
    }

    public void dfs(int i, int j, char[][] grid, int row, int col){
        if(i < 0 || i >= row || j < 0 || j >= col || grid[i][j] != '1')
            return;
        
        grid[i][j] = '2';

        dfs(i-1, j, grid, row, col);
        dfs(i+1, j, grid, row, col);
        dfs(i, j-1, grid, row, col);
        dfs(i, j+1, grid, row, col);
    }
}

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