Hackerrank: a map with Holes

Give you a map of N * n. Each grid in the map has a value indicating the depth of the region. We call a grid in a map empty. if and only when the grid is not at the edge of the map and each grid adjacent to it has a smaller depth than it. Two grids are called adjacent if they have one edge.

You need to find all the holes in the map and useXDescription.

Input Format

The first line contains an integer N, indicating the map size. In the next n rows, each line contains n positive numbers without blank spaces. Each number (1-9) indicates the depth of the corresponding area.

Output Format

Output n rows, indicating the final map result. Each empty space should contain charactersXReplace.

Constraints

1 <= n <= 100

1 ≤ n ≤100

 

Question: A simple simulation.

Question:

 1 import java.io.*; 2 import java.util.*; 3 import java.text.*; 4 import java.math.*; 5 import java.util.regex.*; 6  7 public class Solution { 8     public static void main(String[] args) { 9         Scanner in = new Scanner(System.in);10         int n = in.nextInt();11         int[][] ar = new int[n][n];12         for(int i = 0;i < n;i++){13             String temp = in.next();14             for(int j = 0;j < n;j++)15                 ar[i][j] = temp.charAt(j) - ‘0‘; 16         }17         18         for(int i = 0;i < n;i++){19             StringBuffer sb = new StringBuffer();20             for(int j = 0;j < n;j++){21                 boolean isHole = true;22                 //up23                 if(i-1<0||i+1>=n||j-1<0||j+1>=n)24                     isHole = false;25                 else{26                     if(ar[i-1][j] >= ar[i][j])27                         isHole = false;28                     if(ar[i+1][j] >= ar[i][j])29                         isHole = false;30                     if(ar[i][j-1] >= ar[i][j])31                         isHole = false;32                     if(ar[i][j+1] >= ar[i][j])33                         isHole = false;34                 }35                 36                 if(isHole)37                     sb.append(‘X‘);38                 else {39                     sb.append(ar[i][j]); 40                 }41             }42             System.out.println(sb.toString());43         }44       }45 }