leetcode.463. Island Perimeter

Question

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: 16Explanation: The perimeter is the 16 yellow stripes in the image below:

Solution1

检查四条边，检查每个格子的上下左右

int islandPerimeter(int** grid, int gridRowSize, int gridColSize) {    int i,j,n = 0;    if (grid==NULL)        return 0;        for (i = 0; i < gridRowSize; i++) {        for (j = 0; j < gridColSize; j++) {                        // check rect around line            if (i==0 && grid[i][j] == 1)                n+=1;            if (i==gridRowSize-1 && grid[i][j] == 1)                n+=1;            if (j==0 && grid[i][j] == 1)                n+=1;            if (j==gridColSize-1 && grid[i][j] == 1)                n+=1;                        // check cell's around            if (grid[i][j] == 1 && i-1>=0 && grid[i-1][j] != 1)                n+=1;            if (grid[i][j] == 1 && i+1<gridRowSize && grid[i+1][j] != 1)                n+=1;            if (grid[i][j] == 1 && j+1<gridColSize && grid[i][j+1] != 1)                 n+=1;            if (grid[i][j] == 1 && j-1>=0 && grid[i][j-1] != 1)                 n+=1;                        }    }     return n;}

Solution2

检查横竖2个方向，（别人的最优解）

int islandPerimeter(int** grid, int gridRowSize, int gridColSize) {int i, j, n = 0;// horizontalfor (i = 0; i < gridRowSize; i++) {n += grid[i][0];for (j = 1; j < gridColSize; j++)n += grid[i][j] != grid[i][j - 1];n += grid[i][gridColSize - 1];}// verticalfor (j = 0; j < gridColSize; j++) {n += grid[0][j];for (i = 1; i < gridRowSize; i++)n += grid[i][j] != grid[i-1][j];n += grid[gridRowSize - 1][j];}return n;}