Island Perimeter 海岛周长问题详解

问题详见：Island Perimeter

题目是计算一个用二维矩阵表示的海岛的周长。题目描述如下：
      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:

解题思路：

      由于题目中是一个没有“内湖”的海岛，而且是由一个个相邻的正方形网格组成（虽然实际上的海岛比这复杂很多，但是利用微分思想也还是能解释通）。所以只需要把该网格矩阵中1所代表的陆地的边缘找到即可。这里提供两条思路：1.把每个代表陆地的网格块的4个边相加并减去重复的边；2.找到所有代表海洋的与代表陆地的网格块相邻的边相加。

思路1的算法如下：

class Solution {
public:
    int islandPerimeter(vector<vector<int>>& grid) {
        int ans=0;
        for(int i=0;ifor(int j=0;j0].size();++j){
                if(grid[i][j]==1){
                    ans+=4;
                    if(i!=0&&grid[i-1][j]==1)   ans-=2;
                    if(j!=0&&grid[i][j-1]==1)   ans-=2;
                }
            }
        }
        return ans;
    }
};

其提交运行结果如下：

思路2的算法如下：

class Solution {
public:
    int islandPerimeter(vector<vector<int>>& grid) {
        int ans=0;
        for(int i=0;ifor(int j=0;j0].size();++j){
                if(grid[i][j]==1){
                    if(i==0||grid[i-1][j]==0) ++ans;
                    if(j==0||grid[i][j-1]==0) ++ans;
                    if(i==grid.size()-1||grid[i+1][j]==0) ++ans;
                    if(j==grid[0].size()-1||grid[i][j+1]==0) ++ans;
                }
            }
        }
        return ans;
    }
};

其提交运行结果如下：

结果分析：

      由于思路2的判断和加法运算相对思路1较多，所以尽管两个算法复杂度都是 O(n2) ，但是明显思路1运算时间更快。