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463. Island Perimeter

題目

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:

463. Island Perimeter

分析

求島的周長,其中相鄰的1和0形成一條邊

解答

解法1:(我)異或(185ms)

public class Solution {
    public int islandPerimeter(int[][] grid) {
        int perimeter = 0;
        for (int i = 0; i < grid.length; i++){
            for (int j = 0; j < grid[0].length; j++){
                if (i == 0 && j == 0)//第一行第一列(左上)
                    perimeter += ((grid[i][j] ^ 0) + (grid[i][j] ^ 0));
                else if (i == 0 && j != 0)//第一行(左上)
                    perimeter += ((grid[i][j] ^ 0) + (grid[i][j] ^ grid[i][j-1]));
                else if (j == 0 && i != 0)//第一列(左上)
                    perimeter += ((grid[i][j] ^ grid[i-1][j]) + (grid[i][j] ^ 0));
                else
                    perimeter += ((grid[i][j] ^ grid[i-1][j]) + (grid[i][j] ^ grid[i][j-1]));
                    
                
                if (i == grid.length - 1)//最後一行(下)
                    perimeter += grid[i][j] ^ 0 ;
                if (j == grid[0].length - 1)//最後一列(右)
                    perimeter += grid[i][j] ^ 0;
                
            }
        }
        return perimeter;
    }
}
           

解法2:(我)選出1,計算其上下左右0的個數(155ms√)

public class Solution {
    public int islandPerimeter(int[][] grid) {
        int perimeter = 0;
        for (int i = 0; i < grid.length; i++){
            for (int j = 0; j < grid[0].length; j++){
                if (grid[i][j] == 1){
                    if (i == 0 || grid[i-1][j] == 0){perimeter++;}//up
                    if (j == 0 || grid[i][j-1] == 0){perimeter++;}//left
                    if (i == grid.length - 1 || grid[i+1][j] == 0){perimeter++;}//down
                    if (j == grid[0].length - 1 || grid[i][j+1] == 0){perimeter++;}//right
                }
                
            }
        }
        return perimeter;
    }
}