463. Island Perimeter

Posted 2020-08-31

题目

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:

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分析

求岛的周长，其中相邻的1和0形成一条边

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解答

解法1：（我）异或（185ms）

 1 public class Solution {
 2     public int islandPerimeter(int[][] grid) {
 3         int perimeter = 0;
 4         for (int i = 0; i < grid.length; i++){
 5             for (int j = 0; j < grid[0].length; j++){
 6                 if (i == 0 && j == 0)//第一行第一列（左上）
 7                     perimeter += ((grid[i][j] ^ 0) + (grid[i][j] ^ 0));
 8                 else if (i == 0 && j != 0)//第一行（左上）
 9                     perimeter += ((grid[i][j] ^ 0) + (grid[i][j] ^ grid[i][j-1]));
10                 else if (j == 0 && i != 0)//第一列（左上）
11                     perimeter += ((grid[i][j] ^ grid[i-1][j]) + (grid[i][j] ^ 0));
12                 else
13                     perimeter += ((grid[i][j] ^ grid[i-1][j]) + (grid[i][j] ^ grid[i][j-1]));
14                     
15                 
16                 if (i == grid.length - 1)//最后一行（下）
17                     perimeter += grid[i][j] ^ 0 ;
18                 if (j == grid[0].length - 1)//最后一列（右）
19                     perimeter += grid[i][j] ^ 0;
20                 
21             }
22         }
23         return perimeter;
24     }
25 }

 

解法2：（我）选出1，计算其上下左右0的个数（155ms√）

 1 public class Solution {
 2     public int islandPerimeter(int[][] grid) {
 3         int perimeter = 0;
 4         for (int i = 0; i < grid.length; i++){
 5             for (int j = 0; j < grid[0].length; j++){
 6                 if (grid[i][j] == 1){
 7                     if (i == 0 || grid[i-1][j] == 0){perimeter++;}//up
 8                     if (j == 0 || grid[i][j-1] == 0){perimeter++;}//left
 9                     if (i == grid.length - 1 || grid[i+1][j] == 0){perimeter++;}//down
10                     if (j == grid[0].length - 1 || grid[i][j+1] == 0){perimeter++;}//right
11                 }
12                 
13             }
14         }
15         return perimeter;
16     }
17 }