zoukankan      html  css  js  c++  java
  • [LeetCode]Number of Islands

    原题链接:https://leetcode.com/problems/number-of-islands/

    题意描述:

    Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

    Example 1:

    11110
    11010
    11000
    00000

    Answer: 1

    Example 2:

    11000
    11000
    00100
    00011

    Answer: 3

    题解:

    此题的大意就是找出一个图中的联通区域,一个联通子图便认为是一个岛屿。显然就是对每一个未访问过且是陆地的点进行上下左右搜索,这样即可找到一个联通子图。然后对全图都这样做一遍,便能找到所有的,即结果。代码如下:

     1 public class Solution {
     2     public static boolean flag = false;
     3 
     4     public int numIslands(char[][] grid) {
     5         boolean[][] visited = new boolean[grid.length][];
     6         for (int i = 0; i < grid.length; i++)
     7             visited[i] = new boolean[grid[i].length];
     8         for (int i = 0; i < grid.length; i++) {
     9             for (int j = 0; j < grid[0].length; j++) {
    10                 visited[i][j] = false;
    11             }
    12         }
    13         int res = 0;
    14         for (int i = 0; i < grid.length; i++) {
    15             for (int j = 0; j < grid[0].length; j++) {
    16                 flag = false;
    17                 process(i, j, grid, visited);
    18                 if (flag)
    19                     res++;
    20             }
    21         }
    22         return res;
    23     }
    24 
    25     public void process(int i, int j, char[][] grid, boolean[][] visited) {
    26         if (grid[i][j] != '1' || visited[i][j])
    27             return;
    28         visited[i][j] = true;
    29         flag = true;
    30         if (j >= 1)
    31             process(i, j - 1, grid, visited);
    32         if (i >= 1)
    33             process(i - 1, j, grid, visited);
    34         if(i+1<grid.length)
    35             process(i + 1, j, grid, visited);
    36         if(j+1<grid[0].length)
    37             process(i, j + 1, grid, visited);
    38     }
    39 }
  • 相关阅读:
    关键路径的计算
    JSF简单介绍
    介绍:一款Mathematica的替代开源软件Mathetics
    素材链接
    JSP动作--JSP有三种凝视方式
    【InversionCount 逆序对数 + MergeSort】
    全响应跨设备的Zoomla!逐浪CMS2 x2.0正式公布
    DirectSound的应用
    “海归”首选北上广 薪资期望不太高-有感
    Servlet登陆功能的实现
  • 原文地址:https://www.cnblogs.com/codershell/p/4403489.html
Copyright © 2011-2022 走看看