LeetCode 463: Island Perimeter(島嶼周長)

問題描述：

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.

輸入一個矩陣，1代表陸地，0代表湖泊，求島嶼周長。

Example:

思路一：

分別從上到下，從左到右，若相鄰兩個方塊爲不同屬性（即一個爲陸地，一個爲湖泊）時，島嶼周長加1。主要邊界值的處理，邊界的四周均設爲0即湖泊。使用異或^操作。

Code：

class Solution {
public:
    int islandPerimeter(vector<vector<int>>& grid) {
        int perimeter=0;
        for(int i = 0; i<grid.size();i++){
            int left=0;
            for(int j=0; j<grid[0].size();j++){
                if(left^grid[i][j]) perimeter++;
                left = grid[i][j];
            }
            if(left^0) perimeter++;
        }
        for(int i = 0; i<grid[0].size();i++){
            int left=0;
            for(int j=0; j<grid.size();j++){
                if(left^grid[j][i]) perimeter++;
                left = grid[j][i];
            }
            if(left^0) perimeter++;
        }
        return perimeter;
    }
};

思路二：

若不考慮陸地相鄰，每塊陸地的周長爲4，則島嶼周長爲：陸地塊數*4。考慮到陸地相鄰，則每兩塊相鄰陸地使得島嶼周長減2。遍歷矩陣，統計值爲1的元素即陸地塊數，並查找每塊陸地的左邊和上邊是否爲陸地，若是陸地，則周長減2。注意邊界值的處理。

Code:

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;
    }