leetcode 463. Island Perimeter 周长统计 + 暴力统计即可

浅浅的花香味﹌ 2022-06-02 23:39 167阅读 0赞

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:

![这里写图片描述][SouthEast]

本题很简单，就是统计岛屿的边长，注意本题题意直接计算即可，就是一个很简单的统计即可

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <cmath>
    
    using namespace std;
    
    class Solution 
    {
    public:
        int islandPerimeter(vector<vector<int>>& grid)
        {
            int sum = 0;
            int row = grid.size(), col = grid[0].size();
            for (int i = 0; i < row; i++)
            {
                for (int j = 0; j < col; j++)
                {
                    sum += calaLen(i, j, grid);
                }
            }
            return sum;
        }
    
        int calaLen(int x, int y, vector<vector<int>>& grid)
        {
            int row = grid.size(), col = grid[0].size();
            if (grid[x][y] == 1)
            {
                int c = 0;
                if (x - 1 >= 0 && grid[x - 1][y] == 1)
                    c++;
                if (x + 1 < row && grid[x + 1][y] == 1)
                    c++;
                if (y - 1 >=0 && grid[x][y - 1] == 1)
                    c++;
                if (y + 1 < col && grid[x][y + 1] == 1)
                    c++;
                return 4 - c;
            }
            else
                return 0;
        }
    };

[SouthEast]: /images/20220603/6a0055bb8b9549419b900072dbf0d95f.png