首页 > 代码库 > Island Perimeter Leetcode
Island Perimeter Leetcode
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:
这道题我是按照遍历,看看如果周边有1的话周长就减1的做法做的。
public class Solution { public int islandPerimeter(int[][] grid) { if (grid == null || grid.length == 0 || grid[0].length == 0) { return 0; } int count = 0; for (int i = 0; i < grid.length; i++) { for (int j = 0; j < grid[0].length; j++) { if (grid[i][j] == 1) { count += 4 + isOne(grid, i, j); } } } return count; } private boolean inBound(int[][] grid, int x, int y) { if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) { return false; } return true; } private int isOne(int[][] grid, int x, int y) { int count = 0; int[] dx = {0, 0, 1, -1}; int[] dy = {1, -1, 0, 0}; for (int i = 0; i < 4; i++) { int newX = x + dx[i]; int newY = y + dy[i]; if (inBound(grid, newX, newY) && grid[newX][newY] == 1) { count--; } } return count; } }
做的可能有点复杂了,因为还有更好的做法。只要遍历右边和下边就可以了。最后结果是岛的个数 * 4 - 2 * 右边和下边的个数。因为右边和下边代表有两个岛要减一,所以乘2。
public class Solution { public int islandPerimeter(int[][] grid) { if (grid == null || grid.length == 0 || grid[0].length == 0) { return 0; } int count = 0; 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 < grid.length && grid[i + 1][j] == 1) { count -= 2; } if (j + 1 < grid[0].length && grid[i][j + 1] == 1) { count -= 2; } } } } return count; } }
Island Perimeter Leetcode
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。