zoukankan      html  css  js  c++  java
  • LeetCode 1020. Number of Enclaves

    原题链接在这里:https://leetcode.com/problems/number-of-enclaves/

    题目:

    Given a 2D array A, each cell is 0 (representing sea) or 1 (representing land)

    A move consists of walking from one land square 4-directionally to another land square, or off the boundary of the grid.

    Return the number of land squares in the grid for which we cannot walk off the boundary of the grid in any number of moves.

    Example 1:

    Input: [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
    Output: 3
    Explanation: 
    There are three 1s that are enclosed by 0s, and one 1 that isn't enclosed because its on the boundary.

    Example 2:

    Input: [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
    Output: 0
    Explanation: 
    All 1s are either on the boundary or can reach the boundary.

    Note:

    1. 1 <= A.length <= 500
    2. 1 <= A[i].length <= 500
    3. 0 <= A[i][j] <= 1
    4. All rows have the same size.

    题解:

    For the bondary cell, perform DFS starting from it. Convert all 1 connecting to boundary to 0.

    Iterate A again and count how many cell are still 1.

    Time Complexity: O(m*n). m = A.length. n = A[0].length.

    Space: O(m*n). stack space.

    AC Java:

     1 class Solution {
     2     public int numEnclaves(int[][] A) {
     3         if(A == null || A.length == 0){
     4             return 0;
     5         }
     6         
     7         int m = A.length;
     8         int n = A[0].length;
     9         for(int i = 0; i<m; i++){
    10             dfs(A, i, 0);
    11             dfs(A, i, n-1);
    12 
    13         }
    14         
    15         for(int j = 0; j<n; j++){
    16             dfs(A, 0, j);
    17             dfs(A, m-1, j);
    18         }
    19         
    20         int res = 0;
    21         for(int i = 0; i<m; i++){
    22             for(int j = 0; j<n; j++){
    23                 if(A[i][j] == 1){
    24                     res++;
    25                 }
    26             }
    27         }
    28         
    29         return res;
    30     }
    31     
    32     private void dfs(int [][] A, int i, int j){
    33         if(i<0 || i>=A.length || j<0 || j>=A[0].length || A[i][j]!= 1){
    34             return;
    35         }
    36         
    37         A[i][j] = 0;
    38         dfs(A, i+1, j);
    39         dfs(A, i-1, j);
    40         dfs(A, i, j+1);
    41         dfs(A, i, j-1);
    42     }
    43 }

    类似Surrounded Regions.

  • 相关阅读:
    Use Prerender to improve AngularJS SEO
    Prerender.io
    Prerender Application Level Middleware
    Prerender Application Level Middleware
    正则获取html标签字符串中图片地址
    xml转json
    videojs实现双击视频全屏播放、播放器全屏时视频未全屏
    自己编写jquery插件
    点击回退时需要点击2次才可返回js
    if中有逗号的写法
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11941498.html
Copyright © 2011-2022 走看看