LeetCode_463. Island Perimeter

463. Island Perimeter
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:

题意理解：计算岛屿周长—（0，1）（1，0）部分。

一看到题目的立马反应是求相邻元素相减不为0的边长和，这个用Matlab有diff函数实现相邻元素求差，C++的话，只想到遍历(遍历有点容易晕)。
遍历分为3部分，第一部分是边缘元素，第二部分是0到1的跳变，第三部分是1到0的跳变。

c++:

class Solution{public:    int islandPerimeter(vector<vector<int>>& grid){        int num=0;        for(int i=0;i<grid.size();i++)        {            for(int j=0;j<grid[0].size();j++)            {                if(grid[i][j])                {                    if(i==0||grid[i-1][j]==0)   num++;  //水平方向：0->1                    if(i==grid.size()-1||grid[i+1][j]==0)   num++; //1->0                    if(j==0||grid[i][j-1]==0)   num++; // 竖直方向：0->1                    if(j==grid[0].size()-1||grid[i][j+1]==0)   num++; //1->0                    }            }        }        return num;    }};

reference:
http://blog.csdn.net/jmspan/article/details/53254264

Top Solution
思路：先统计有多少个1的块，每个方块有4条边，再除去不符合条件的边，即相邻两块均为1的边。

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