【leetcode】【Easy】【463. Island Perimeter】

problem link:https://leetcode.com/problems/island-perimeter/

code1:效率较高的两种解法，打败一半的人

//查看右边和下边的邻居

public class Solution {
    public int islandPerimeter(int[][] grid) {
        int islands = 0, neighbours = 0;

        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                if (grid[i][j] == 1) {
                    islands++; // count islands
                    if (i < grid.length - 1 && grid[i + 1][j] == 1) neighbours++; // count down neighbours
                    if (j < grid[i].length - 1 && grid[i][j + 1] == 1) neighbours++; // count right neighbours
                }
            }
        }

        return islands * 4 - neighbours * 2;
    }
}

//与上面解法思想上是一样的，只是考虑的是左边和上边的邻居

public int islandPerimeter(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) return 0;
        int result = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    result += 4;
                    if (i > 0 && grid[i-1][j] == 1) result -= 2;
                    if (j > 0 && grid[i][j-1] == 1) result -= 2;
                }
            }
        }
        return result;
    }

code2:效率非常低的一种解法：直接遍历数

public class Solution {
    public int islandPerimeter(int[][] grid) {
		int rowLen = grid.length;
		int colLen = grid[0].length;

		int res = 0;
		for (int i = 0; i < rowLen - 1; i++) {
			for (int j = 0; j < colLen - 1; j++) {
				// res += (grid[i][j] ^ grid[i][j + 1] + grid[i][j] ^ grid[i +
				// 1][j]);
				//这样会有错误！
				res += grid[i][j] ^ grid[i][j + 1];
				res += grid[i][j] ^ grid[i + 1][j];
			}
		}

		for (int i = 0; i < rowLen; i++) {
			res += grid[i][0] ^ 0;
			res += grid[i][colLen - 1] ^ 0;
			if (i != (rowLen - 1))
				res += grid[i][colLen - 1] ^ grid[i + 1][colLen - 1];
		}

		for (int i = 0; i < colLen; i++) {
			res += grid[0][i] ^ 0;
			res += grid[rowLen - 1][i] ^ 0;
			if (i != (colLen - 1))
				res += grid[rowLen - 1][i] ^ grid[rowLen - 1][i + 1];
		}

		return res;
	
    }
}