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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:

首先是自己写的一个low比算法，主要是对二维数组的每个点都判断其是否为 1，若为1，则接着判断该元素是否为边界元素，以及其周围元素是否为0元素，依次来决定是否加上边长。

int count_subPer(vector
   
    
     >& res,int i,int j){    int count = 0;    if (res[i][j] == 1){    //对于每个点的以及周围点的判断写的可能有点乱，没有做优化        if (i == 0)count++;        if (i == res.size() - 1)count++;        if (j == 0)count++;        if (j == res[0].size() - 1)count++;        if (i - 1 >= 0 && res[i - 1][j] == 0)count++;        if (i + 1 < res.size() && res[i + 1][j] == 0)count++;        if (j - 1 >= 0 && res[i][j - 1] == 0)count++;        if (j + 1 < res[0].size() && res[i][j + 1] == 0)count++;    }    return count;}int islandPerimeter(vector
     
      
       >& grid) {    if (grid.size() == 0)return 0;    int sum = 0;    for (int i = 0; i < grid.size(); i++){        for (int j = 0; j < grid[0].size(); j++){            if (grid[i][j] == 0)continue;            sum += count_subPer(grid, i, j);        }    }    return sum;}
      
     
    
   

第二种方法：

int islandPerimeter(vector
   
    
     >& grid) {    if (grid.size() == 0)return 0;    int cnt = 0, repeat = 0;    for (int i = 0; i < grid.size(); i++){        for (int j = 0; j < grid[0].size(); j++){            if (!grid[i][j])continue;            cnt++;            if (i != 0 && grid[i - 1][j])repeat++;            if (j != 0 && grid[i][j - 1])repeat++;        }    }    return cnt * 4 - repeat * 2;}