如何将一个数组分成3个部分,每个部分的总和大致相等

时间:2011-11-28 07:48:14

标签: c# algorithm

我有一个排列的数组,我想将它分成3个部分,以便它们的总和彼此最接近。

Ex: I have this array:

    10, 8, 8, 7, 6, 6, 6, 5

so it'll be divided into 3 part like:

    p1 {10,8} sum = 18
    p2 {8,7,6} sum = 21
    p3 {6,6,5} sum = 17

6 个答案:

答案 0 :(得分:9)

原始海报已经有了一个工作解决方案(在评论中注明),将数组分成两部分,数额相等;请拨打此split2。可以使用split2构建三部分版本。

  1. 向数组添加一个等于原始数字总和的三分之一的新数字。
  2. 使用split2将数组拆分为两部分。
  3. 一部分有添加的数字;删除它。
  4. 使用split2将其他部分拆分为两部分。

答案 1 :(得分:2)

这就像two-Partition问题,它是NP-Hard,但不是强烈意义上的,你可以有一个O(nK)算法,其中K是输入和的大小,参见pseudo polynomial time algorithm for subset sum,另请参阅我对divide-list-in-two-parts-that-their-sum-closest-to-each-other的回答,但在您的情况下,您应该添加另一个维度来处理它。

答案 2 :(得分:1)

尝试以下代码

int total = 0, partSum = 0, partIndex = 0;
int noOfParts = 3; //Initialize the no. of parts
int[] input = { 10, 8, 8, 7, 6, 6, 6, 5 };
int[] result = new int[noOfParts]; //Initialize result array with no. of locations equal to no. of parts, to store partSums
foreach (int i in input) //Calculate the total of input array values
{
    total += i;
}
int threshold = (total / noOfParts) - (total / input.Length) / 2; //Calculate a minimum threshold value for partSum
for (int j = input.Length - 1; j > -1; j--)
{
    partSum += input[j]; //Add array values to partSum incrementally
    if (partSum >= threshold) //If partSum reaches the threshold value, add it to result[] and reset partSum  
    {
        result[partIndex] = partSum;
        partIndex += 1;
        partSum = 0;
        continue;
    }
}
if (partIndex < noOfParts) //If no. of parts in result[] is less than the no. of parts required, add the remaining partSum value
{
    result[partIndex] = partSum;
}
Array.Reverse(result);
foreach (int k in result)
{
    Console.WriteLine(k);
}
Console.Read();     

我已经使用数组中的各种值(按降序排列)测试了这个值,并且使用了不同的值。零件(3,4,5 ......)并取得了良好的效果。

答案 3 :(得分:1)

// calculate total
total = 0;
for(i = 0; i != size; ++i) {
   total += array[i];
}

// partition
n_partitions = 3;
current_partition = 1;
subtotal = array[0];
for(i = 1; i != size; ++i) {
   if(subtotal + array[i] > total / n_partitions) {
      // start new partition;
      current_partition++;
      subtotal = array[i];
   } else {
      // push to current partition
      subtotal += array[i];
   }
}

答案 4 :(得分:0)

更新了代码:

我建议的方法如下(代码如下):

  • 创建数据结构(集合等)以表示您需要的输出部分的数量(在您的示例3中)
  • 按降序对输入数组进行排序。
  • 遍历输入数组的元素并为每个值:
    • 选择一个输出部分以将值放入(这应该是当前具有最低总和的输出部分..)
    • 将值添加到选定的输出部分

根据上述逻辑,您将始终添加具有最低总值的输出部分(这将有助于保持相似整体值的部分)。

(在下面的代码示例中,我已跳过数组排序步骤,因为您的示例已经排序)

代码:

        // the input array
        int[] inputArray = new int[] { 10, 8, 8, 7, 6, 6, 6, 5 };

        // the number of parts you want
        int numberOfOutputParts = 3;

        // create the part structures
        List<Part> listOfParts = new List<Part>();

        for(int i =0; i < numberOfOutputParts; i++)
        {
            listOfParts.Add(new Part());
        }

        // iterate through each input value
        foreach (int value in inputArray)
        {
            // find the part with the lowest sum
            int? lowestSumFoundSoFar = null;
            Part lowestValuePartSoFar = null;

            foreach(Part partToCheck in listOfParts)
            {
                if (lowestSumFoundSoFar == null || partToCheck.CurrentSum < lowestSumFoundSoFar)
                {
                    lowestSumFoundSoFar = partToCheck.CurrentSum;
                    lowestValuePartSoFar = partToCheck;
                }
            }

            // add the value to that Part
            lowestValuePartSoFar.AddValue(value);
        }

上面使用的Part类的代码(尽管你可以使用更好的东西如下):

public class Part
{
    public List<int> Values
    {
        get;
        set;
    }

    public int CurrentSum
    {
        get;
        set;
    }

    /// <summary>
    /// Default Constructpr
    /// </summary>
    public Part()
    {
        Values = new List<int>();
    }

    public void AddValue(int value)
    {
        Values.Add(value);
        CurrentSum += value;
    }
}

答案 5 :(得分:0)

你能试试我的样本,这可能对你有所帮助

我的算法: 1 /通过输出数组的数量计算数组的平均值(exp:值= 3)

2 /对数组求和,直到Sum的最小间隙与平均值相比(以1 /计算)

3 /执行步骤2,直到您到达数组编号的末尾

我使用C#3.5进行测试

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Collections;

namespace WindowsFormsApplication2
{
    public partial class Form2 : Form
    {
        public Form2()
        {
            InitializeComponent();
        }

        ArrayList inputValue = new ArrayList();
        int avgValue = 0;
        bool isFinish = false;
        private void button1_Click(object sender, EventArgs e)
        {
            #region Init data
            isFinish = false;
            avgValue = 0;
            inputValue.Clear();
            listBox1.Items.Clear();
            //assum you input valid number without space and in desc sorting order 
            string[] arrNumber = textBox1.Text.Split(new char[] { ',' }, StringSplitOptions.RemoveEmptyEntries);
            int numberOfBreak = 3;
            int record = Convert.ToInt32(arrNumber[0]);//update the record with the maximum value of the array numbers
            for (int i = 0; i < arrNumber.Length; i++)
            {
                inputValue.Add(Convert.ToInt32(arrNumber[i]));
            }

            foreach (object obj in inputValue)
            {
                avgValue += (int)obj;
            }
            avgValue = avgValue / numberOfBreak;
            #endregion
            int lastIndex = 0;
            while (!isFinish)
            {
                int index = GetIndex(lastIndex);
                string sResult = "";
                for (int i = lastIndex; i <= index; i++)
                {
                    sResult += inputValue[i].ToString() + "-";
                }
                listBox1.Items.Add(sResult);
                if (index + 1 < inputValue.Count)
                {
                    lastIndex = index + 1;
                }
                sResult = "";
            }
        }

        private int GetIndex(int startIndex)
        {
            int index = -1;
            int gap1 = Math.Abs(avgValue - (int)inputValue[startIndex]);
            int tempSum = (int)inputValue[startIndex];
            if (startIndex < inputValue.Count - 1)
            {

                int gap2 = 0;
                while (gap1 > gap2 && !isFinish)
                {
                    for (int i = startIndex + 1; i < inputValue.Count; i++)
                    {
                        tempSum += (int)inputValue[i];

                        gap2 = Math.Abs(avgValue - tempSum);
                        if (gap2 <= gap1)
                        {
                            gap1 = gap2;
                            gap2 = 0;
                            index = i;
                            if (startIndex <= inputValue.Count - 1)
                            {
                                startIndex += 1;
                            }
                            else
                            {
                                isFinish = true;
                            }
                            if (startIndex == inputValue.Count - 1)
                            {
                                index = startIndex;
                                isFinish = true;
                            }
                            break;
                        }
                        else
                        {
                            index = i - 1;
                            break;
                        }
                    }
                }


            }
            else if (startIndex == inputValue.Count - 1)
            {
                index = startIndex;
                isFinish = true;
            }
            else
            {
                isFinish = true;
            }
            return index;
        }
    }
}