【LeetCode】Island Perimeter 解题报告

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[LeetCode]

https://leetcode.com/problems/island-perimeter/

• Total Accepted: 16143
• Total Submissions: 28552
• Difficulty: Easy

Question

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:

Ways

有点类似小学火柴棍的问题。我最初的想法是统计一的个数，找规律，试验了几次之后发现貌似是1的个数*2+2，但是这个是错误的。比如最角落的1如果被周围的1所包围的话，实际上是不贡献边的。

后来想到，应该把边角的被包围的1去掉，中间的被四个1包围的1也给去掉，剩余的1的个数*4.这样很麻烦，因为判断处于边角位置的1就要写4个If,判断被包围的1也要判断四次。

看到top的答案我才明白，可以统计1的个数*4 再减去 与其有交集的1的个数*2.为了防止减去的边不被重复计算，所以，只判断右边和下边的1.代码如下。

public class Solution {

    public int islandPerimeter(int[][] grid) {

        int count=0;

        int neighbours=0;

        for(int i=0; i<grid.length; i++){

            for(int j=0 ;j<grid[i].length; j++){

                if(grid[i][j]==1){

                    count++;// count islands

                    if(j<grid[i].length-1 && grid[i][j+1]==1)

                        neighbours++;// count right neighbours

                    if(i<grid.length-1 && grid[i+1][j]==1)

                        neighbours++;// count down neighbours

                }

            }

        }

        return 4*count-2*neighbours;

    }

}

果然这种方法判断的次数要少一点。

Date

2017 年 1 月 6 日