463. Island Perimeter

最新推荐文章于 2022-10-28 18:08:21 发布

wchstrife

最新推荐文章于 2022-10-28 18:08:21 发布

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本文介绍了一种计算二维网格中岛屿周长的方法。该算法通过遍历数组，对每个标记为岛屿的单元格进行检查，计算其周围相邻的岛屿数量来确定最终周长。

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463. Island Perimeter

Description Hints Submissions Solutions

Total Accepted: 41559
Total Submissions: 72872
Difficulty: Easy
Contributors:amankaraj

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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题意：

用一个二维数组表示一个图，1代表岛屿，一个岛屿有4个边，与其他岛屿相邻的部分不算边，并且岛内没有空隙，求岛的边长

算法思路：

遍历二维数组，如果元素为1，边数+4，再去寻找这个岛屿周边为1的岛屿，如果有就边数-1

代码：

public class Solution {
	public static int islandPerimeter(int[][] grid) {
		
        int m = grid.length;
		
		int n = grid[0].length;
		
		int count = 0;
		
		for(int i=0; i<m; i++){
			for(int j=0; j<n; j++){
				if(grid[i][j] == 1){
					count += 4;
					if(i>0 && grid[i-1][j]==1){
						count--;
					}
					if(j>0 && grid[i][j-1]==1){
						count--;
					}
					if(i<m-1 && grid[i+1][j] == 1){
						count--;
					}
					if(j<n-1 && grid[i][j+1] == 1){
						count--;
					}
				}				
				
			}					
		}
		return count;
    }
	
    public static void main(String[] args){
    	int[][] a= {{0,1,0,0},
    			 {1,1,1,0},
    			 {0,1,0,0},
    			 {1,1,0,0}};
		System.out.println(islandPerimeter(a));
	}
}