leetcode.463. Island Perimeter

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本文介绍了一种计算二维网格中岛屿周长的方法。该方法适用于由1（陆地）和0（水域）组成的矩阵，其中岛屿定义为至少一个陆地单元格，并且不包含内部水域。文章提供了两种算法实现：一种通过检查每个单元格的边界和相邻单元格；另一种通过检查水平和垂直方向的变化。

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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: 16
Explanation: 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;
	// horizontal
	for (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];
	}
	// vertical
	for (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;
}