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C#教程之C#归并排序的实现方法(递归,非递归,自然
//Main:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Merge
{
class Program
{
static void Main(string[] args)
{
while (true)
{
Console.WriteLine("请选择：");
Console.WriteLine("1.归并排序（非递归）");
Console.WriteLine("2.归并排序（递归）");
Console.WriteLine("3.归并排序（自然合并）");
Console.WriteLine("4.退出");
int Arraynum = Convert.ToInt32(Console.ReadLine());
switch (Arraynum)
{
case 4:
Environment.Exit(0);
break;
case 1:
Console.WriteLine("Please Input Array Length");
int Leng271 = Convert.ToInt32(Console.ReadLine());
Function obj1 = new Function(Leng271);
Console.WriteLine("The original sequence:");
Console.WriteLine(obj1);
Console.WriteLine("'MergeSort' Finaly Sorting Result:");
obj1.ToMergeSort();
Console.WriteLine(obj1);
break;
case 2:
Console.WriteLine("Please Input Array Length");
int Leng272 = Convert.ToInt32(Console.ReadLine());
Function obj2 = new Function(Leng272);
Console.WriteLine("The original sequence:");
Console.WriteLine(obj2);
Console.WriteLine("'RecursiveMergeSort' Finaly Sorting Result:");
obj2.ToRecursiveMergeSort();
Console.WriteLine(obj2);
break;
case 3:
Console.WriteLine("Please Input Array Length");
int Leng273 = Convert.ToInt32(Console.ReadLine());
Function obj3 = new Function(Leng273);
Console.WriteLine("The original sequence:");
Console.WriteLine(obj3);
obj3.ToNaturalMergeSort();
Console.WriteLine();Console.WriteLine();
Console.WriteLine("'NaturalMergeSort' Finaly Sorting Result:");
Console.WriteLine(obj3);
break;
}
}
}
}
}
//Class:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Merge
{
// 【example 2.7】//抱歉，实在不知怎么学习英语，语法什么错误之处还请见谅。
class Function
{
private int Groups;
private int CopyGroups;
private int mergerows;
private int[] Array27;
private static Random ran = new Random();
public Function(int length)
{
Array27 = new int[length];
for (int i = 0; i < length; i++)
Array27[i] = /*Convert.ToInt32(Console.ReadLine()); //*/ran.Next(1, 100);
}
//选择
public void ToMergeSort()
{
MergeSort(Array27);
}
public void ToRecursiveMergeSort()
{
RecursiveMergeSort(Array27, 0, Array27.Length - 1);
}
public void ToNaturalMergeSort()
{
NaturalMergeSort(Array27);
}
/// <summary>
/// 归并排序(递归）
/// 核心思想：（分治）
/// 将待排序元素（递归直至元素个数为１）分成左右两个大小大致相同的2个子集合，然后，
/// 分别对2个子集合进行排序，最终将排好序的子集合合并成为所要求的排好序的集合．
/// 核心算法时间复杂度：
/// T(n)=O(nlogn)
/// 参考 优秀代码: http://zh.wikipedia.org/wiki/%E5%90%88%E4%BD%B5%E6%8E%92%E5%BA%8F
/// http://www.cnblogs.com/mingmingruyuedlut/archive/2011/08/18/2144984.html
/// </summary>
/// <param name="Array"></param>
/// <param name="left"></param>
/// <param name="right"></param>
public void RecursiveMergeSort(int[] Array, int left, int right)
{
int middle = (left + right) / 2;
if (left < right)
{
//对前半部分递归拆分
RecursiveMergeSort(Array, left, middle);
//对后半部分递归拆分
RecursiveMergeSort(Array, middle + 1, right);
MergeOne(Array, left, middle, right);
}
}
public void MergeOne(int[] Array, int left, int middle, int right)
{
int leftindex = left;
int rightindex = middle + 1;
//动态临时二维数组存放分割为两个小Array的数组排列顺序后的数据
int[] merge = new int[right + 1];
int index = 0;
//对两个小数组合并排序
while (leftindex <= middle && rightindex <= right)
merge[index++] = (Array[leftindex] - Array[rightindex]) >= 0 ? Array[rightindex++] : Array[leftindex++];
//有一侧子数列遍历完后，将另外一侧子数列剩下的数依次放入暂存数组中（有序）
if (leftindex <= middle)
{
for (int i = leftindex; i <= middle; i++)
merge[index++] = Array[i];
}
if (rightindex <= right)
{
for (int i = rightindex; i <= right; i++)
merge[index++] = Array[i];
}
//将有序的数列 写入目标数组 ，即原来把Array数组分为两个小Array数组后重新有序组合起来(覆盖原数组）
index = 0;
for (int i = left; i <= right; i++)
Array[i] = merge[index++];
}
/// <summary>
/// 归并排序(非递归）
/// 核心思想：（分治）
/// 对n个数的数列每相邻两个元素排序，组成n/2个或（n+1)/2个子数组，单个的不比了直接进入下一轮。
/// 然后对每个相邻的子数组再排序，以此类推最后得到排好序的数列
/// forexample： 59 35 54 28 52
/// 排序And分： 35 59. 28 54. 52
/// 排序And分： 28 35 54 59. 52
/// 结果： 28 35 52 54 59
/// 核心算法时间复杂度：
/// T(n)=O(nlogn)
/// </summary>
/// <param name="Array"></param>
public void MergeSort(int[] Array)
{
//index固定的数组
int[] merge = new int[Array.Length];
int P = 0;
while (true)
{
int index = 0;
//子数组元素的个数
int ENumb = (int)Math.Pow(2, P);
//一个子数组中的元素个数与数组的一半元素个数比较大小
//最糟糕的情况最右边的数组只有一个元素
if (ENumb < Array.Length)
{
while (true)
{
int TorFAndrightindex = index;
//最后一个子数组的第一个元素的index与数组index相比较
if (TorFAndrightindex <= Array.Length - 1)
MergeTwo(Array, merge, index, ENumb);
else
break;
index += 2 * ENumb;
}
}
else
break;
P++;
}
}
public void MergeTwo(int[] Array, int[] merge, int index, int ENumb)
{
//换分两个子数组的index(千万不能用middle = (right + left) / 2划分）
// 1
int left = index;
int middle = left + ENumb - 1;
//（奇数时）
//排除middleindex越界
if (middle >= Array.Length)
{
middle = index;
}
//同步化merge数组的index
int mergeindex = index;
// 2
int right;
int middleTwo = (index + ENumb - 1) + 1;
right = index + ENumb + ENumb - 1;
//排除最后一个子数组的index越界.
if (right >= Array.Length - 1)
{
right = Array.Length - 1;
}
//排序两个子数组并复制到merge数组
while (left <= middle && middleTwo <= right)
{
merge[mergeindex++] = Array[left] >= Array[middleTwo] ? Array[middleTwo++] : Array[left++];
}
//两个子数组中其中一个比较完了（Array[middleTwo++] 或Array[left++]），
//把其中一个数组中剩下的元素复制进merge数组。
if (left <= middle)
{
//排除空元素.
while (left <= middle && mergeindex < merge.Length)
merge[mergeindex++] = Array[left++];
}
if (middleTwo <= right)
{
while (middleTwo <= right)
merge[mergeindex++] = Array[middleTwo++];
}
//判断是否合并至最后一个子数组了
if (right + 1 >= Array.Length)
Copy(Array, merge);
}
/// <summary>
/// 自然归并排序:
/// 对于初始给定的数组,通常存在多个长度大于1的已自然排好序的子数组段.
/// 例如,若数组a中元素为{4,8,3,7,1,5,6,2},则自然排好序的子数组段
/// 有{4,8},{3,7},{1,5,6},{2}.
/// 用一次对数组a的线性扫描就足以找出所有这些排好序的子数组段.
/// 然后将相邻的排好序的子数组段两两合并,
/// 构成更大的排好序的子数组段({3,4,7,8},{1,2,5,6}).
/// 继续合并相邻排好序的子数组段,直至整个数组已排好序.
/// 核心算法时间复杂度：
/// T(n)=○(n);
/// </summary>
public void NaturalMergeSort(int[] Array)
{
//得到自然划分后的数组的index组（每行为一个自然子数组）
int[,] PointsSymbol = LinearPoints(Array);
//子数组只有一个。
if (PointsSymbol[0, 1] == Array.Length - 1)
return;
//多个(至少两个子数组）...
else
//可以堆栈调用吗？
NaturalMerge(Array, PointsSymbol);
}
public void NaturalMerge(int[] Array, int[,] PointsSymbol)
{
int left;
int right;
int leftend;
int rightend;
mergerows = GNumberTwo(Groups);
CopyGroups = Groups;
//固定状态
int[] TempArray = new int[Array.Length];
//循环取每个自然子数组的index
while (true)
{
// int Temprow = 1;
//只记录合并后的子数组(”《应该为》“动态的）
int[,] TempPointsSymbol = new int[mergerows, 2];
int row = 0;
do
{
//最重要的判断：最后(一组子数组)是否可配对
if (row != CopyGroups - 1)
{ //以上条件也可以含有(& 和&&的区别)短路运算
//参考:http://msdn.microsoft.com/zh-cn/library/2a723cdk(VS.80).aspx
left = PointsSymbol[row, 0];
leftend = PointsSymbol[row, 1];
right = PointsSymbol[row + 1, 0];
rightend = PointsSymbol[row + 1, 1];
MergeThree(Array, TempArray, left, leftend, right, rightend);
MergePointSymbol(PointsSymbol, TempPointsSymbol, row);
}
else
{
////默认剩下的单独一个子数组已经虚拟合并。然后Copy进TempArray。
int TempRow = PointsSymbol[row, 0];
int TempCol = PointsSymbol[row, 1];
while (TempRow <= TempCol)
TempArray[TempRow] = Array[TempRow++];
//TempPointSymbol完整同步
TempPointsSymbol[row / 2, 0] = PointsSymbol[row, 0];
TempPointsSymbol[row / 2, 1] = PointsSymbol[row, 1];
break;//重新开始新一轮循环。
}
row += 2;
// Temprow++;
//合并到只有一个子数组时结束循环
if (TempPointsSymbol[0, 1] == Array.Length - 1)
break;
}//判断别进入越界循环(可以进孤单循环）这里指的是PointsSymbol的子数组个数
while (row <= CopyGroups - 1);
//
Copy(Array, TempArray);
//更新子数组index，row为跳出循环的条件（最后单个子数组或下一个越界的第一个）
UpdatePointSymbol(PointsSymbol, TempPointsSymbol, row);
//改变TempPointsSymbol的行数（合并后子数组数）
mergerows = GNumber(mergerows);
CopyGroups = GNumberTwo(CopyGroups);
//合并到只有一个子数组时结束循环
if (PointsSymbol[0, 1] == Array.Length - 1)
break;
}
//输出
}
public int GNumber(int Value)
{
if (Value % 2 == 0)
Value /= 2;
else
Value -= 1;
return Value;
}
public int GNumberTwo(int Value)
{
if (Value % 2 == 0)
mergerows = Value / 2;
else
mergerows = Value / 2 + 1;
return mergerows;
}
public void MergeThree(int[] Array, int[] Temp, int left, int leftend, int right, int rightend)
{
//合并语句
int index = left;
while (left <= leftend && right <= rightend)
Temp[index++] = Array[left] >= Array[right] ? Array[right++] : Array[left++];
while (left <= leftend)
Temp[index++] = Array[left++];
while (right <= rightend)
Temp[index++] = Array[right++];
}
public void MergePointSymbol(int[,] PointsSymbol, int[,] TempPointsSymbol, int row)
{
int rowindex = row / 2;
TempPointsSymbol[rowindex, 0] = PointsSymbol[row, 0];
TempPointsSymbol[rowindex, 1] = PointsSymbol[row + 1, 1];
}
public void UpdatePointSymbol(int[,] PointsSymbol, int[,] TempPointsSymbol, int rows)
{
int row = 0;
//if (mergerows % 2 == 0)
//{
for (; row < TempPointsSymbol.GetLength(0); row++)
{
for (int col = 0; col < 2; col++)
PointsSymbol[row, col] = TempPointsSymbol[row, col];
}
//后面的清零
for (; row < PointsSymbol.GetLength(0); row++)
{
for (int col2 = 0; col2 < 2; col2++)
PointsSymbol[row, col2] = 0;
}
//}
////补剩下的index组，
//else
//{
// for (int row2 = 0; row2 < TempPointsSymbol.GetLength(0); row2++)
// {
// for (int col3 = 0; col3 < 2; col3++)
// PointsSymbol[row2, col3] = TempPointsSymbol[row2, col3];
// }
// //最后一个子数组的index只有一个。
// int row3 = TempPointsSymbol.GetLength(0);
// PointsSymbol[row3, 0] = PointsSymbol[rows, 0];
// PointsSymbol[row3, 1] = PointsSymbol[rows, 1];
// //后面的清零
// for (int row4 = row3 + 1; row4 < PointsSymbol.GetLength(0); row4++)
// {
// for (int col4 = 0; col4 < 2; col4++)
// PointsSymbol[row4, col4] = 0;
// }
//}
}
public int[,] LinearPoints(int[] Array)
{
Groups = 1;
int StartPoint = 0;
int row = 0;
int col = 0;
//最糟糕的情况就是有Array.Length行。
int[,] PointsSet = new int[Array.Length, 2];
//线性扫描Array，划分数组
//初始前index=0
PointsSet[row, col] = 0;
do
{
//判断升序子数组最终的index开关
bool Judge = false;
//从Array第二个数判断是否要结束或者是否是升序子数组.
while (++StartPoint < Array.Length && Array[StartPoint] < Array[StartPoint - 1])
{
//打开第一个升序子数组结束的index开关
Judge = true;
//重新开始第二个升序子数组的前index
PointsSet[row, col + 1] = StartPoint - 1;
//计算子数组个数
Groups++;
//换行记录自然子数组的index
row++;
break;
//--StartPoint;
}
//升序子数组结束index
if (Judge)
PointsSet[row, col] = StartPoint;
//else
// --StartPoint;
} while (StartPoint < Array.Length);
//最终index=StartPoint - 1，但是糟糕情况下还有剩余若干行为： 0,0 ...
PointsSet[row, col + 1] = StartPoint - 1;
//调用展示方法
DisplaySubarray(Array, PointsSet, Groups);
return PointsSet;
}
public void DisplaySubarray(int[] Array, int[,] PointsSet, int Groups)
{
Console.WriteLine("Subarray is {0}:", Groups);
//展示子数组的前后index
for (int r = 0; r < Groups; r++)
{
for (int c = 0; c < PointsSet.GetLength(1); c++)
{
Console.Write(PointsSet[r, c]);
if (c < PointsSet.GetLength(1) - 1)
Console.Write(",");
}
Console.Write("\t\t");
}
Console.WriteLine();
//展示分出的子数组
for (int v = 0; v < Groups; v++)
{
int i = 1;
for (int r = PointsSet[v, 0]; r <= PointsSet[v, 1]; r++)
{
Console.Write(Array[r] + " ");
i++;
}
if (i <= 3)
Console.Write("\t\t");
else
Console.Write("\t");
if (PointsSet[v, 1] == Array.Length)
break;
}
}
public void Copy(int[] Array, int[] merge)
{
//一部分排好序的元素替换掉原来Array中的元素
for (int i = 0; i < Array.Length; i++)
{
Array[i] = merge[i];
}
}
//输出
public override string ToString()
{
string temporary = string.Empty;
foreach (var element in Array27)
temporary += element + " ";
temporary += "\n";
return temporary;
}
}
}
