zoukankan      html  css  js  c++  java
  • [leetcode]694. Number of Distinct Islands你究竟有几个异小岛?

    Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

    Count the number of distinct islands. An island is considered to be the same as another if and only if one island can be translated (and not rotated or reflected) to equal the other.

    Example 1:

    11000
    11000
    00011
    00011
    

    Given the above grid map, return 1.

    Example 2:

    11011
    10000
    00001
    11011

    Given the above grid map, return 3.

    Notice that:

    11
    1
    

    and

     1
    11
    

    are considered different island shapes, because we do not consider reflection / rotation.

    Note: The length of each dimension in the given grid does not exceed 50.

    题目

    此题是[leetcode]200. Number of Islands岛屿个数小岛题的变种

    要数出不同小岛的个数,题意定义了不同小岛: 互相不能通过非反射、旋转而转化

    思路

    - In 2D array, once we find 1st '1',  we find an island.  Meanwhile, we set a 'draw_begin' , ready to draw such island's shape

    - DFS to loop through 4 different directions,  checking whether another '1' is connected with.  Meanwhile, we use u, d, l, r to draw island's shape.

    - Once finish DFS, we set a 'draw_end',  indicating such island's drawing is done

    - Why we use 'draw_begin' and  'draw_end' to set a bound ? 

    -  要去handle 类似以下这种情况

    11     and     1   
    1            1 1

    - Visited spots won't be visited again because they are updated to '0'

    代码

     1 class Solution {
     2    public int numDistinctIslands(int[][] grid) {
     3         //put different drawing shape represented by string to hashset
     4         Set<String> set = new HashSet<>();
     5         for(int i = 0; i < grid.length; i++) { 
     6             for(int j = 0; j < grid[i].length; j++) {
     7                 if(grid[i][j] != 0) {
     8                     StringBuilder sb = new StringBuilder();
     9                     dfs(grid, i, j, sb, "draw_begin"); // set a bound
    10                     set.add(sb.toString());
    11                 }
    12             }
    13         }
    14        // how many diffent shapes represented by string in hashset  
    15         return set.size();
    16     }
    17     private void dfs(int[][] grid, int i, int j, StringBuilder sb, String dir) {
    18         // out of bound or not an island 
    19         if(i < 0 || i == grid.length || j < 0 || j == grid[i].length 
    20            || grid[i][j] == 0) return;
    21         
    22         sb.append(dir);
    23         grid[i][j] = 0;// mark a visted spot 
    24         dfs(grid, i-1, j, sb, "u");// up
    25         dfs(grid, i+1, j, sb, "d");// down 
    26         dfs(grid, i, j-1, sb, "l");// left
    27         dfs(grid, i, j+1, sb, "r");// right
    28         sb.append("draw_end"); // set a bound 
    29     }
    30 }
  • 相关阅读:
    该死的兼容性
    Tip:解决DesignMode不能正确反应是否处于设计模式的问题
    Oh, ListView里竟然什么也不显示,也不报错!
    Tip: Dock is better than Anchor
    Tip: Asp.net下载默认文件名里包含空格时,如何防止FireFox只截取空格前一部分作为文件名
    c#的DateTime.Now函数详解
    C#中的委托和事件(初稿)
    C#多线程学习
    .NET(c#)new关键字的三种用法
    Silverlight – WCF – MaxItemsInObjectGraph
  • 原文地址:https://www.cnblogs.com/liuliu5151/p/9825367.html
Copyright © 2011-2022 走看看