LeetCode - 463. Island Perimeter - O(MN)- (C++)

本文介绍了一个算法问题，即计算一个二维网格中岛屿的周长。该岛屿由相邻的陆地方格组成，且不包含内部湖泊。通过检查每个陆地方格的四周来确定其对总周长的贡献。

原题

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:

给定一个二维整数的方格地图，1代表陆地，0代表有水的地方。每个方格都是垂直或者水平连接的，没有斜线。被水包围的中间有一个小岛（小岛可能包含了很多个的陆地方格），小岛里没有岛中湖（内部的水和外部的水是连通的）。每个方格单元的边长都是1。各个地方都是成直角的，长度宽度不超过100，判断小岛的边长。

思路

判断其四面是否有陆地，四面都有陆地的对周长无贡献，两面有陆地的贡献周长为2，一面有陆地的贡献周长为3，四面都没有陆地的贡献周长为4。

代码

class Solution {
public:
    int islandPerimeter(vector<vector<int>>& grid) {
        int sum=0;
        int m=grid.size(),n=grid[0].size();
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
                if(grid[i][j]){ 
                if(i==0||grid[i-1][j]==0) sum++;
                if(i==m-1||grid[i+1][j]==0) sum++;
                if(j==0||grid[i][j-1]==0) sum++;
                if(j==n-1||grid[i][j+1]==0) sum++;
                }
        return sum;
    }
};

转载于:https://www.cnblogs.com/rgvb178/p/6121274.html