LeetCode-Island Perimeter

本文介绍了一种计算二维网格中岛屿周长的方法。在一个由1和0组成的矩阵中,1代表岛屿,0代表海水,通过遍历网格并计算每个岛屿与相邻岛屿的关系,最终得出岛屿的总周长。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Description:
You are given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water. Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells). The island doesn’t have “lakes” (water inside that isn’t connected to the water around the island). One cell is a square with side length 1. The grid is rectangular, width and height don’t exceed 100. Determine the perimeter of the island.

Example:

[[0,1,0,0],
 [1,1,1,0],
 [0,1,0,0],
 [1,1,0,0]]

Answer: 16
Explanation: The perimeter is the 16 yellow stripes in the image below:
在这里插入图片描述

题意:在一个网格中,用1代表一个岛屿,用0代表海水,现在要求计算所有的岛屿的周长;

解法:对于每个岛屿来说,其周长为4,每相邻另外一个岛屿其周长减1,因此,我们遍历整个网格中所有的岛屿,在其初始周长4的前提下,减去相邻的岛屿数,最后对有的的岛屿的周长累加得到结果;

Java
class Solution {
    public int islandPerimeter(int[][] grid) {
        int result = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                if (grid[i][j] == 1) {
                    result += 4 - aroundCell(grid, i, j);
                }
            }
        }
        return result;
    }
    
    private int aroundCell(int[][] grid, int row, int col) {
        int cnt = 0;
        if (row - 1 >= 0 && grid[row - 1][col] == 1) cnt++;
        if (col - 1 >= 0 && grid[row][col - 1] == 1) cnt++;
        if (row + 1 < grid.length && grid[row + 1][col] == 1) cnt++;
        if (col + 1 < grid[row].length && grid[row][col + 1] == 1) cnt++;
        return cnt;
    }
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值