简单版数独计算器-升级版

只能算初级的..高级的就溢出了


就算内存无穷大.可能性超过了20亿就数组放不下了



因为是广度优先吧..所以..争取能写个深度优先的办法


哎..好难啊..头发掉了好多

package com.ansj.ansjIndex;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;

public class CopyOfShuDu {
	public static void main(String[] args) {
		int[][] ints = {{4,7,0,6,0,0,0,0,0},
				{0,0,1,9,4,0,0,0,0},
				{0,9,6,5,1,7,2,4,0},
				{0,0,2,0,9,0,0,3,0},
				{0,0,0,1,3,2,0,0,0},
				{0,8,0,0,5,0,6,0,0},
				{0,3,9,2,8,5,7,6,0},
				{0,0,0,0,6,1,3,0,0},
				{0,0,0,0,0,9,0,5,8}};

		add(ints);

	}

	public static void add(int[][] ints) {
		int[][][] intsTemp = new int[9][9][0];
		for (int i = 0; i < ints.length; i++) {
			for (int j = 0; j < ints[i].length; j++) {
				if (ints[i][j] == 0) {
					int[] tempInt = havePosb(ints, i, j) ;
					if(tempInt==null){
						add(ints) ;
						return ;
					}
					intsTemp[i][j] = tempInt;
				}
			}
		}

		if(isOk(ints)){
			for (int i = 0; i < ints.length; i++) {
				printArray(ints[i]) ;
			}
			System.out.println("--------------------------------------------------------");
			return ;
		}
		
		zuhe(ints, intsTemp);
		
		for (int i = 0; i < list.size(); i++) {
			fillVacancy(ints,list.get(i)) ;
			
			if(isOk(ints)){
				for (int j = 0; j < ints.length; j++) {
					printArray(ints[j]) ;
				}
				System.out.println("--------------------------------------------------------");
			}else{
			}
		}
		
		


	}
	
	public static void fillVacancy(int[][] ints , int[] temp){
		int count = 0 ;
		for (int i = 0; i < ints.length; i++) {
			for (int j = 0; j < ints[i].length ; j++) {
				if(ints[i][j]<=0){
					ints[i][j] = temp[count]*-1 ;
					count++ ;
				}
			}
		}
		
	}
	
	//广度优先排列组合
	public static void zuhe(int[][] ints, int[][][] intsTemp) {
		List<int[]> listArray = new ArrayList<int[]>() ;
		for (int i = 0; i < intsTemp.length; i++) {
			for (int j = 0; j < intsTemp[i].length; j++) {
				if(intsTemp[i][j].length>0){
					listArray.add(intsTemp[i][j]) ;
				}
			}
		}
		
		addHead(listArray) ;
	}
	
	private static LinkedList<int[]> list= new LinkedList<int[]>();
	private static int pamerLength ;
	
	public static void addHead(List<int[]> listArray){
		int[] temp = listArray.get(0) ;
		for (int i = 0; i < temp.length; i++) {
			list.add(Arrays.copyOfRange( temp,i , i+1)) ;
		}
		pamerLength = listArray.size() ;
		addHeap(listArray) ;
	}
	
	public static void addHeap(List<int[]> listArray){
		int innerLength = 1;   
        while (true) {
            int[] temp = list.pollFirst();   
            innerLength = temp.length ;   
            // 深度够了则执行完成   
            if (innerLength == pamerLength) {   
                list.addFirst(temp);   
                return;   
            } else {   
                for (int innerInt : listArray.get(innerLength)) {   
                    // 依次入队列的末端  
                	int[] newTemp = new int[innerLength+1] ;
                	for (int i = 0; i < temp.length; i++) {
						newTemp[i] = temp[i] ;
					}
                	newTemp[temp.length] = innerInt ;
                    list.addLast(newTemp);   
                }   
            }   
        }  
	}
	
	
	
	/**
	 * 找出数组中空白处可以填写的数字
	 * 
	 * @param ints
	 * @param i
	 * @param j
	 */
	public static int[] havePosb(int[][] ints, int i, int j) {
		int[] arr = new int[10];
		for (int k = 0; k < 9; k++) {
			arr[Math.abs(ints[k][j])] = 1;
		}

		for (int k = 0; k < 9; k++) {
			arr[Math.abs(ints[i][k])] = 1;
		}

		int n = i / 3;
		int m = j / 3;

		for (int k = 1; k < 3; k++) {
			arr[Math.abs(ints[n * 3 + k][m * 3 + k])] = 1;
		}
		List<Integer> all = new ArrayList<Integer>();
		for (int k = 1; k < arr.length; k++) {
			if (arr[k] == 0) {
				all.add(k);
			}
		}
		int[] tempInt = new int[all.size()];

		for (int k = 0; k < tempInt.length; k++) {
			tempInt[k] = all.get(k);
		}
		
		if(tempInt.length==1){
			ints[i][j] = tempInt[0] ;
			tempInt = null ;
		}
		
		return tempInt;
	}

	public static boolean isOk(int[][] ints) {
		for (int i = 0; i < ints.length; i++) {
			for (int j = 0; j < ints.length; j++) {
				if (ints[i][j] == 0) {
					return false;
				}
			}
		}

		
		
		for (int i = 0; i < ints.length; i++) {
			for (int j = 0; j < ints.length; j++) {
				if(ints[i][j]>0){
					continue ;
				}
				
//				for (int j1 = 0; j1 < ints.length; j1++) {
//					printArray(ints[j1]) ;
//				}
//				System.out.println("--------------------------------------------------------");
				
				int[] arr = new int[9];
				
				for (int k = 0; k < 9; k++) {
					arr[Math.abs(ints[i][k])-1]++ ; 
				}
				for (int k = 0; k < 9; k++) {
					arr[Math.abs(ints[k][j])-1]++ ; 
				}
//				printArray(arr) ;
				
				for (int k = 0; k < arr.length; k++) {
					if(arr[k]!=2)return false ;
				}
			}
		}

		return true;
	}

	/**
	 * 重置数组
	 * 
	 * @param ints
	 */
	private static void resetArray(int[][] ints) {
		for (int i = 0; i < ints.length; i++) {
			for (int j = 0; j < ints[i].length; j++) {
				if (ints[i][j] < 0) {
					ints[i][j] = 0;
				}
			}
		}
	}

	private static void printArray(int[] ints) {
		for (int i = 0; i < ints.length; i++) {
			System.out.print(ints[i] + "	");
		}
		System.out.println();
	}

}

评论

3 楼 ansjsun 2012-02-10  

awaterway 写道

能否告知，写这个大概花了多长时间呢？

忘记了些的有问题哈。。估计一天就搞定吧。。这个不是很复杂。。用到了个排列组合就是。。如果没记错的话

2 楼 awaterway 2012-02-10  

能否告知，写这个大概花了多长时间呢？

1 楼 awaterway 2012-02-10  

要我写就早变成光头了。。。。