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  • Interview How to Count Squares

    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 }
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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/5174194.html
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