109、岛屿的周长

题目描述

首先想到的是暴力法求解，即每次看到有1时就看其四周有没有0或者是边界，如果有那么将周长加1
代码：

class Solution {
    public int islandPerimeter(int[][] grid) {
        int count = 0;
        for (int i = 0; i < grid.length; i++) {
			for (int j = 0; j < grid[i].length; j++) {
				int is = grid[i][j];
				if(is == 1){
					if(i == 0 || grid[i-1][j] == 0){
						count++;
					}
					if(j == 0 || grid[i][j-1] == 0){
						count++;
					}
					if(i == grid.length -1 || grid[i+1][j] == 0){
						count++;
					}
					if(j == grid[i].length - 1 || grid[i][j+1] == 0){
						count++;
					}
				}	
			}	
		}
        return count;
    }
}

排名靠前的代码
其实思路差不多的

class Solution {
    public int islandPerimeter(int[][] grid) {
        int count = 0;//record the value of the answer
        for(int i = 0;i < grid.length;i++){
            for(int j = 0;j < grid[0].length;j++){
                if(grid[i][j] == 1){
                    count += 4;
                    if(i - 1 >= 0 && grid[i - 1][j] == 1){
                        count -= 2;
                    }
                    if(j - 1 >= 0 && grid[i][j - 1] == 1){
                        count -= 2;
                    }
                }

            }
        }
        return count;
    }
}