Interview How to Count Squares

Posted 2020-06-12

火柴拼出多少个正方形 http://matchstickpuzzles.blogspot.com/2011/06/55-4x4-square-how-many-squares.html

输入是两个二维数组ver 和 hor, 若是有火柴就是1, 没有就是0.

dpHor 表示横方向上有多少连续火柴，dpVer表示纵方向上有多少连续火柴。

最后以左上角第一根横着的火柴为根基检查是否能组成正方形。

Time Complexity: O(n^3). Space: O(n^2).

 1 import java.util.*;
 2 public class countSquare{
 3     public static void main(String [] args){
 4         int [][] hor = {{1,1},{1,0},{1,1}};
 5         int [][] ver = {{1,1,1},{1,1,1}};
 6         System.out.println("Number of square: " + countSquare(hor,ver));
 7     }
 8 
 9     private static int countSquare(int [][] hor, int [][] ver){
10         if(hor == null || ver == null || hor.length == 0 || ver.length == 0 || hor[0].length == 0 || ver[0].length == 0){
11             return 0;
12         }
13 
14         int [][] dpHor = new int[hor.length][hor[0].length];
15         int [][] dpVer = new int[ver.length][ver[0].length];
16 
17         for(int i = 0; i<hor.length; i++){
18             for(int j = 0; j<hor[0].length; j++){
19                 if(hor[i][j] == 1){
20                     dpHor[i][j] = j == 0 ? 1 : dpHor[i][j-1] + 1;
21                 }else{
22                     dpHor[i][j] = 0;
23                 }
24             }
25         }
26 
27         for(int j = 0; j<ver[0].length; j++){
28             for(int i = 0; i<ver.length; i++){
29                 if(ver[i][j] == 1){
30                     dpVer[i][j] = i == 0 ? 1 : dpVer[i-1][j] + 1;
31                 }else{
32                     dpVer[i][j] = 0;
33                 }
34             }
35         }
36 
37         System.out.println("dpHor is " + Arrays.deepToString(dpHor));
38         System.out.println("dpVer is " + Arrays.deepToString(dpVer));
39 
40         int res = 0;
41         for(int i = 0; i<hor.length; i++){
42             for(int j = 0; j<hor[0].length; j++){
43                 for(int len = 1; len<= Math.min(ver.length-i, hor[0].length-j); len++){
44                     if(dpHor[i][j+len-1] >= len && dpHor[i+len][j+len-1] >= len && dpVer[i+len-1][j] >= len && dpVer[i+len-1][j+len] >= len){
45                         res++;
46                         System.out.println("i = " + i + ", j = " + j + ", len = " + len + ", res = " + res);
47                     }
48                 }
49             }
50         }
51         return res;
52     }
53 }