zoukankan      html  css  js  c++  java
  • [LeetCode] 463. Island Perimeter

    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:

    Input:
    [[0,1,0,0],
     [1,1,1,0],
     [0,1,0,0],
     [1,1,0,0]]
    
    Output: 16
    
    Explanation: The perimeter is the 16 yellow stripes in the image below:
    
    

    岛屿的周长。题意是给一个二维矩阵,只有0和1。1表示岛屿,0表示水。请返回岛屿的周长。这个题虽然是number of islands的followup,但是其实跟BFS或者DFS没什么关系。思路是遍历input,如果当前坐标的值是1,说明是岛屿,周长 * 4;同时判断当前坐标的右边一个位置和下面一个位置是否也是岛屿,如果是,则减去两条边,因为两个邻居各自都失去一条边。

    时间O(n^2)

    空间O(1)

    Java实现

     1 class Solution {
     2     public int islandPerimeter(int[][] grid) {
     3         int islands = 0;
     4         int neighbors = 0;
     5         for (int i = 0; i < grid.length; i++) {
     6             for (int j = 0; j < grid[0].length; j++) {
     7                 if (grid[i][j] == 1) {
     8                     islands++;
     9                     if (i < grid.length - 1 && grid[i + 1][j] == 1) {
    10                         neighbors++;
    11                     }
    12                     if (j < grid[0].length - 1 && grid[i][j + 1] == 1) {
    13                         neighbors++;
    14                     }
    15                 }
    16             }
    17         }
    18         return islands * 4 - neighbors * 2;
    19     }
    20 }

    JavaScript实现

     1 /**
     2  * @param {number[][]} grid
     3  * @return {number}
     4  */
     5 var islandPerimeter = function (grid) {
     6     let m = grid.length;
     7     let n = grid[0].length;
     8     let islands = 0;
     9     let neighbors = 0;
    10     for (let i = 0; i < m; i++) {
    11         for (let j = 0; j < n; j++) {
    12             if (grid[i][j] == 1) {
    13                 islands++;
    14                 if (i < m - 1 && grid[i + 1][j] == 1) {
    15                     neighbors++;
    16                 }
    17                 if (j < n - 1 && grid[i][j + 1] == 1) {
    18                     neighbors++;
    19                 }
    20             }
    21         }
    22     }
    23     return islands * 4 - neighbors * 2;
    24 };

    LeetCode 题目总结

  • 相关阅读:
    Chat Icon
    docker 容器无root 权限,如何获得docker容器里面的root权限
    yolo训练自己的数据
    jetson nano(1-1) 系统烧录和备份
    jetson nano(2)软件环境开发
    jetson nano(1-2)配置VNC
    坐标映射(remap重映射)
    opencv图像格式
    matlab相机标定导出xml文件
    jupyter安装和链接aconda
  • 原文地址:https://www.cnblogs.com/cnoodle/p/12850382.html
Copyright © 2011-2022 走看看