leetcode[Island Perimeter]//待整理多种解法

解法一：

public class Solution {boolean up(int[][] grid, int x, int y){//用于测试坐标(x,y)处的上方是否有岛，即是否没有公共边界，是否可以有边界x = x - 1;//x才是控制行数的,y是控制列数的/*if(y < 0 || grid[x][y] != 1){return false;}else{return true;}*/return x < 0 || grid[x][y] != 1;//比起注释部分更简单的写法}boolean left(int[][] grid, int x, int y){y = y - 1;return y < 0 || grid[x][y] != 1;}boolean down(int[][] grid, int x, int y){x = x + 1;return x > grid.length - 1 || grid[x][y] != 1;//逻辑或，左边为真就不计算右边了，保证不会有越界的问题}boolean right(int[][] grid, int x, int y){y = y + 1;//System.out.println(x);//System.out.println(grid[0].length - 1);return y > grid[0].length - 1 || grid[x][y] != 1;}    public int islandPerimeter(int[][] grid) {        int perimeter = 0;//perimeter代表周长            for(int i = 0; i < grid.length; i++){        for(int j = 0; j < grid[i].length; j++){        //System.out.println(i + "  " + j);        if(grid[i][j] == 1){//建立在(i,j)本身是岛的基础上            if(up(grid, i,j)) perimeter++;            if(left(grid, i,j)) perimeter++;            if(down(grid, i,j)) perimeter++;            if(right(grid, i,j)) perimeter++;        }        }        }                return perimeter;    }}