Leetcode-463. Island Perimeter

最新推荐文章于 2025-09-12 18:08:49 发布

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本文介绍了一个计算二维网格中岛屿周长的问题解决方案。通过遍历找到第一个陆地单元格，并使用递归方法计算整个岛屿的周长。文章提供了一个Java实现示例。

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发优快云，mcf171专栏。

博客链接：mcf171的博客

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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:

这个题目没啥好说的，因为知道一定有一块陆地，所以用遍历的方式找到第一个1，然后用启发的方式就行了。 Your runtime beats 31.98% of java submissions.

public class Solution {
    public int islandPerimeter(int[][] grid) {
        if(grid.length == 0) return 0;
        int length = 0;
        for(int i = 0; i < grid.length; i ++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == 1){
                    length = active(grid,i,j);break;
                }
            }
            if(length != 0) break;
        }
        return length;
    }

    public int active(int[][] grid, int i, int j){
        int[] x = { i + 1, i - 1};
        int[] y = { j + 1, j - 1};
        int length = 0;
        grid[i][j] = 2;

        if( (i - 0) * ( i - grid.length + 1) <= 0 && (j - 0) * ( j - grid[0].length + 1) <= 0){
            int start = 0, end = 2;
            if(i == 0) {length ++ ; end = 1;}
            if( i == grid.length - 1) {length ++ ;start = 1;}

            for(int k = start ; k < end ; k ++){
                int item = x[k];
                if(grid[item][j] == 0) length ++;
                else if(grid[item][j] == 1) length += active(grid,item,j);
            }
            start = 0; end = 2;
            if(j == 0) {length ++ ; end = 1;}
            if(j == grid[0].length - 1) {length ++; start = 1;}

            for(int k = start ; k < end ; k ++){
                int item = y[k];
                if(grid[i][item] == 0)length ++;
                else if(grid[i][item] == 1) length += active(grid, i, item);
            }
        }

        return length;
    }
}