利用Java快速查找21位花朵数示例代码

前言

本文主要给大家介绍了关于利用Java快速查找21位花朵数的相关内容，分享出来供大家参考学习，下面话不多说了，来一起看看详细的介绍吧。

以前备赛的时候遇到的算法题，求所有21位花朵数，分享一下，供大家参考，效率已经很高了。

示例代码

package com.jianggujin;

import java.math.BigInteger;
import java.util.ArrayList;
import java.util.List;


public class NarcissusNumber
{
 
 private BigInteger[] powerOf10;
 
 private BigInteger[][] preTable1;
 
 private int[][] preTable2;
 
 private int[] selected = new int[10];
 
 private int length;

 
 private List results;
 
 private int numberSystem = 10;

 
 private NarcissusNumber(int n)
 {
  powerOf10 = new BigInteger[n + 1];
  powerOf10[0] = BigInteger.ONE;
  length = n;
  results = new ArrayList();

  // 初始化powerPowerOf10
  for (int i = 1; i <= n; i++)
  {
   powerOf10[i] = powerOf10[i - 1].multiply(BigInteger.TEN);
  }

  preTable1 = new BigInteger[numberSystem][n + 1];
  preTable2 = new int[numberSystem][n + 1];

  // preTable[i][j] 0-i的N次方出现0-j次的值
  for (int i = 0; i < numberSystem; i++)
  {
   for (int j = 0; j <= n; j++)
   {
   preTable1[i][j] = new BigInteger(new Integer(i).toString()).pow(n)
     .multiply(new BigInteger(new Integer(j).toString()));

   for (int k = n; k >= 0; k--)
   {
    if (powerOf10[k].compareTo(preTable1[i][j]) < 0)
    {
     preTable2[i][j] = k;
     break;
    }
   }
   }
  }
 }

 public static List search(int num)
 {
  NarcissusNumber narcissusNumber = new NarcissusNumber(num);
  narcissusNumber.search(narcissusNumber.numberSystem - 1, BigInteger.ZERO, narcissusNumber.length);
  return narcissusNumber.getResults();
 }

 
 private void search(int currentIndex, BigInteger sum, int remainCount)
 {
  if (sum.compareTo(powerOf10[length]) >= 0)
  {
   return;
  }

  if (remainCount == 0)
  {
   // 没数可选时
   if (sum.compareTo(powerOf10[length - 1]) > 0 && check(sum))
   {
   results.add(sum);
   }
   return;
  }

  if (!preCheck(currentIndex, sum, remainCount))
  {
   return;
  }

  if (sum.add(preTable1[currentIndex][remainCount]).compareTo(powerOf10[length - 1]) < 0)// 见结束条件2
  {
   return;
  }

  if (currentIndex == 0)
  {
   // 选到0这个数时的处理
   selected[0] = remainCount;
   search(-1, sum, 0);
  }
  else
  {
   for (int i = 0; i <= remainCount; i++)
   {
   // 穷举所选数可能出现的情况
   selected[currentIndex] = i;
   search(currentIndex - 1, sum.add(preTable1[currentIndex][i]), remainCount - i);
   }
  }
  // 到这里说明所选数currentIndex的所有情况都遍历了
  selected[currentIndex] = 0;
 }

 
 private boolean preCheck(int currentIndex, BigInteger sum, int remainCount)
 {
  if (sum.compareTo(preTable1[currentIndex][remainCount]) < 0)// 判断当前值是否小于PreTable中对应元素的值
  {
   return true;// 说明还有很多数没选
  }
  BigInteger max = sum.add(preTable1[currentIndex][remainCount]);// 当前情况的最大值
  max = max.divide(powerOf10[preTable2[currentIndex][remainCount]]);// 取前面一部分比较
  sum = sum.divide(powerOf10[preTable2[currentIndex][remainCount]]);

  while (!max.equals(sum))
  {
   // 检验sum和max首部是否有相同的部分
   max = max.divide(BigInteger.TEN);
   sum = sum.divide(BigInteger.TEN);
  }

  if (max.equals(BigInteger.ZERO))// 无相同部分
  {
   return true;
  }

  int[] counter = getCounter(max);

  for (int i = 9; i > currentIndex; i--)
  {
   if (counter[i] > selected[i])// 见结束条件3
   {
   return false;
   }
  }
  for (int i = 0; i <= currentIndex; i++)
  {
   remainCount -= counter[i];
  }
  return remainCount >= 0;// 见结束条件4
 }

 
 private boolean check(BigInteger sum)
 {
  int[] counter = getCounter(sum);
  for (int i = 0; i < numberSystem; i++)
  {
   if (selected[i] != counter[i])
   {
   return false;
   }
  }
  return true;
 }

 
 private int[] getCounter(BigInteger value)
 {
  int[] counter = new int[numberSystem];
  char[] sumChar = value.toString().toCharArray();

  for (int i = 0; i < sumChar.length; i++)
  {
   counter[sumChar[i] - '0']++;
  }

  return counter;
 }

 
 public List getResults()
 {
  return results;
 }

 public static void main(String[] args)
 {
  int num = 21;
  System.err.println("正在求解" + num + "位花朵数");
  long time = System.nanoTime();
  List results = NarcissusNumber.search(num);
  time = System.nanoTime() - time;
  System.err.println("求解时间:t" + time / 1000000000.0 + "s");
  System.err.println("求解结果:t" + results);
 }
}

运行查看结果：

正在求解21位花朵数

求解时间: 0.327537257s

求解结果: [128468643043731391252, 449177399146038697307]

总结

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