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

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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;    }}

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