排序算法C Part01

1、冒泡排序

//冒泡排序BubbleSort
//每次扫描,比较和交换相邻元素,保证一次扫描后,最大元素在最后一个
//每次扫描后,扫描队列长度减一
//时间复杂度:O(N^2)
//空间复杂度:O(N)
//稳定性:稳定
//输入:要排序的数组,数组长度
//输出:void,Q[]将处于排序状态
void BubbleSort(int Q[], int N)
{
  int i;
  int swap=1;
  int Temp;
  
  while(swap>0)
    {
      swap=0;
      for(i=1;i<N;i++)
	{
	  if(Q[i-1]>Q[i])
	    {
	      swap++;
	      Temp=Q[i-1];
	      Q[i-1]=Q[i];
	      Q[i]=Temp;
	    }
	}
    }
}

2、选择排序

//选择排序SelectionSort
//从头扫描到尾,每次将最小元素放到第一个
//每次扫描后,队列长度减一
//时间复杂度:O(N^2)
//空间复杂度:O(N)
//稳定性:稳定
//输入:要排序的数组,数组长度
//输出:void,Q[]将处于排序状态
void SelectionSort(int Q[], int N)
{
  int i,j,k;
  int Temp;
  for(i=0;i<N;i++)
    {
      k=i;
      for(j=i+1;j<N;j++)
	{
	  if(Q[j]<Q[i])k=j;
	}  

      Temp=Q[k];
      Q[k]=Q[i];
      Q[i]=Temp;
    }
}

3、插入排序

//插入排序InsertionSort
//起始队列长度为1,每次队列长度加1
//通过元素移动,将队列调整为有序状态
//时间复杂度:O(N^2)
//空间复杂度:O(N)
//稳定性:稳定
//输入:要排序的数组,数组长度
//输出:void,Q[]将处于排序状态
void InsertionSort(int Q[], int N)
{
  int i,j;
  int Temp;
  for(i=1;i<N;i++)
    {
      Temp = Q[i];
      for(j=i;j>0 && Q[j-1]>Temp;j--)
	{
	  Q[j]=Q[j-1];
	}

      Q[j]=Temp;
    }
}

4、希尔排序

//希尔排序ShellSort
//指定一个步长,将需要排序的数字,按序号%步长分组
//对每组进行插入排序,然后减小步长,再次分组,排序直到步长为1结束
//时间复杂度:O(n^2)
//空间复杂度:O(n)
//稳定性:稳定
//输入:要排序的数组,数组长度
//输出:void,Q[]将处于排序状态
void ShellSort(int Q[], int N)
{
  int i,j;
  int Temp;
  int Step;

  for(Step=N/2;Step>0;Step=Step/2)
    {
      for(i=Step;i<N;i++)
	{
	  Temp = Q[i];
	  for(j=i;j>=Step && Q[j-Step]>Temp;j=j-Step)
	    {
	      Q[j]=Q[j-Step];
	    }
	  Q[j]=Temp;
	}
      //dumpArray(Q,N);
    }
}

5、归并排序

//归并排序MergeSort
//二路归并首先将归并长度定为2,在归并集内进行排序
//然后归并长度×2,将有序集合进行归并
//时间复杂度:O(NlogN)
//空间复杂度:O(N)
//稳定性:稳定
//输入:要排序的数组,数组长度
//输出:void,Q[]将处于排序状态
void Merge(int Q[],int TempArray[], int left, int mid, int right)
{
  int i=left,j=mid,k=left;

  while(i<mid && j<=right)
    {
      if(Q[i]>Q[j])
	{
	  TempArray[k++]=Q[j++];
	}
      else
	{
	  TempArray[k++]=Q[i++];
	}
    }

  while(i<mid)
    {
	  TempArray[k++]=Q[i++];
    }

  while(j<=right)
    {
      	  TempArray[k++]=Q[j++];
    }

  for(k=left;k<=right;k++)
    {
      Q[k]=TempArray[k];
    }

}

void MSort(int Q[],int TempArray[], int left, int right)
{
  int center;

  if(left<right)
    {
      center=(left+right)/2;
      MSort(Q,TempArray,left,center);
      MSort(Q,TempArray,center+1,right);
      Merge(Q,TempArray,left,center+1,right);
    }
}

void MergeSort(int Q[], int N)
{
  int *TempArray = malloc(N*sizeof(int));

  if(TempArray!=NULL)
    {
      MSort(Q,TempArray,0,N-1);
    }
  else
    {
      //Todo: deal with not enough memory
      //RaiseError("MergeSort: not enough memory");
    }
}

八后问题Ruby02

Queen.rb

class Queen
  
  def initialize()
    @v=0
  end
  
  def canSetQueen(lst,s,x,y)
      lstNo = Array.new()
    
      for i in 0..s-1 do
        p = Pt.new(x,i)
        lstNo.push(p)
      end
      
      for j in 0..s-1 do
        p = Pt.new(j,y)
        lstNo.push(p)
      end 
      
      x0=x
      y0=y
      while x0<s and y0<s
        p=Pt.new(x0,y0)
        lstNo.push(p)
        x0=x0+1
        y0=y0+1
      end
      
      x0=x
      y0=y
      while x0>=0 and y0>=0
        p=Pt.new(x0,y0)
        lstNo.push(p)
        x0=x0-1
        y0=y0-1
      end
      
      x0=x
      y0=y
      while x0>=0 and y0<s
        p=Pt.new(x0,y0)
        lstNo.push(p)
        x0=x0-1
        y0=y0+1
      end
      
      x0=x
      y0=y
      while x0<s and y0>=0
        p=Pt.new(x0,y0)
        lstNo.push(p)
        x0=x0+1
        y0=y0-1
      end
      
      while(lstNo.length>0)
        p=lstNo.pop
        if(p.y==lst[p.x])
          return false
        end
      end
      
      return true
  end
    
  def findQueen(lst,s,x) 
      
      if(x>=s)
        @v=@v+1
        puts("]>>solution No."+@v.to_s)
        pp lst
        return
      end
      
      for y in 0..s-1 do
        if(canSetQueen(lst,s,x,y))
          lst.push(y)   
          findQueen(lst,s,x+1)
          lst.pop
        end
      end
       
    end
end

class Pt
  attr_accessor:x
  attr_accessor:y
  def initialize(x,y)
      @x=x
      @y=y
  end
end

test.rb

require "pp"
require "./Queen.rb"

len = 8
lst = Array.new()

q=Queen1.new
q.findQueen(lst,len,0)

puts("end")

八后问题Ruby01

Queen.rb

class Queen
  
  def initialize()
    @v=0
  end
  
  def arrCopy2(arr)
    arr0=Array.new(arr.length)
    for i in 0..arr.length-1 do
      arr0[i]=Array.new(arr[i])
    end
    
    return arr0
  end
  
  def addQueen(arr,x,y)
    setQueen(arr,x,y,1)
  end
  
  def setQueen(arr,x,y,n)
    for i in 0..arr.length-1 do
      arr[x][i]=n
    end
    
    for j in 0..arr[x].length-1 do
      arr[j][y]=n
    end 
    
    x0=x
    y0=y
    while x0<arr.length and y0<arr&#91;x0&#93;.length
      arr&#91;x0&#93;&#91;y0&#93;=n
      x0=x0+1
      y0=y0+1
    end
    
    x0=x
    y0=y
    while x0>=0 and y0>=0
      arr[x0][y0]=n
      x0=x0-1
      y0=y0-1
    end
    
    x0=x
    y0=y
    while x0>=0 and y0<arr&#91;x0&#93;.length
      arr&#91;x0&#93;&#91;y0&#93;=n
      x0=x0-1
      y0=y0+1
    end
    
    x0=x
    y0=y
    while x0<arr.length and y0>=0
      arr[x0][y0]=n
      x0=x0+1
      y0=y0-1
    end
    
    return arr
  end
  
  def findQueen(arr,lst,x) 
    
    if(x>=arr.length)
      @v=@v+1
      puts("]>>solution No."+@v.to_s)
      pp lst
      return
    end
    
    for y in 0..arr[x].length-1 do
      if(arr[x][y]==0)
        lst.push(y)
        arr0=arrCopy2(arr)
        addQueen(arr,x,y)   
        findQueen(arr,lst,x+1)
        arr=arr0
        lst.pop
      end
    end
     
  end
  
end

test.rb

require "pp"
require "./Queen.rb"

len = 8
arr=Array.new(len){Array.new(len, 0)}
#fuck
#arr = Array.new(len, Array.new(len,0))

lst = Array.new()

q=Queen.new
q.findQueen(arr,lst,0)

puts("end")

排序算法Ruby Part2

6、Heap sort 堆排序

  # Heap sort 堆排序
  # 使用堆进行排序,每次重建堆后,将堆顶元素放到堆数组最后,
  # 堆数组长度减一
  # Time Complexity: О(nlogn) average and worst-case
  # Space Complexity: О(n) total, O(1) auxiliary
  # Stable: Yes
  def heapsort(container)
    return container if container.size <= 1
    
    buildheap(container)
    size = container.size
    while size > 0
      container[0], container[size-1] = container[size-1], container[0]
      size = size - 1
      heapify(container, 0, size)
    end
    
    return container
  end
  
  private
  
  # 重建堆
  # 某节点及其所有子节点,符合最大堆的特性
  def heapify(container, idx, size)
    left_idx = 2 * idx + 1
    right_idx = 2 * idx + 2
    bigger_idx = idx
    bigger_idx = left_idx if left_idx < size && container[left_idx] > container[idx]
    bigger_idx = right_idx if right_idx < size && container[right_idx] > container[bigger_idx]
    if bigger_idx != idx
      container[idx], container[bigger_idx] = container[bigger_idx], container[idx]
      heapify(container, bigger_idx, size)
    end
    return container
  end
  
  # 初始化最大堆,堆顶元素为container[0]
  # 元素i的左节点为2*i+1,元素i的右节点为2*i+2
  def buildheap(container)
    last_parent_idx = container.length / 2 - 1
    i = last_parent_idx
    while i >= 0
      heapify(container, i, container.size)
      i = i - 1
    end
  end

7、Quicksort 快速排序

  # Quicksort 快速排序
  # 用分治法进行排序,充分利用了每次比较的结果,
  # 每次排序都将排序对象分成了两组,分别进行排序
  # Time Complexity: О(n log n) average, O(n^2) worst-case
  # Space Complexity: О(n) auxiliary
  # Stable: No
  def quicksort(container)
    return container if container.size<2

    i=0
    j=container.size-1
    key=container[i]

    while(i<j) do
      while(i<j and container[j]>key) do
        j=j-1
      end
      if(i<j)
        container[i]=container[j]
        i=i+1
      end

      while(i<j and container[i]<key) do
        i=i+1
      end
      if(i<j)
        container[j]=container[i]
        j=j-1
      end
    end
    container[i]=key
    
    f= 0<=i ? quicksort(container[0,i+1]): nil
    t= i<=container.size-1 ? quicksort(container[i+1,container.size-1]): nil

    return t if(f==nil)
    return f if(t==nil)
    return f+t
  end

8、Counting Sort 计数排序

  # Counting Sort 计数排序
  # 计算最小值和最大值,利用标志位来标记元素出现次数
  # Time Complexity: О(n+k) for best, average and worst-case
  # Space Complexity: О(n+k) auxiliary
  # Stable: Yes
  def countingsort(container)
    min = container.min
    max = container.max
    counts = Array.new(max-min+1, 0)

    container.each do |n|
      counts[n-min] += 1
    end

    j=0
    for i in 0..counts.size-1 do
      if(counts[i]!=0)
        while(counts[i]>0) do
          container[j] = min+i
          counts[i]-=1
          j+=1
        end
      end

      i+=1
    end

    return container
  end

9、Radix Sort 基数排序

  # Radix Sort 基数排序
  # 从个位数进行进行排序,一直到最高位
  # Time Complexity: О(k*n) for worst-case
  # Space Complexity: О(k+n) for worst-case
  # Stable: Yes
  def radixsort(container)
    
    max = container.max
    d = Math.log10(max).floor + 1

    (1..d).each do |i|
      radix = []
      (0..9).each do |j|
        radix[j] = []
      end

      container.each do |n|
        kth = calckth(n, i)
        radix[kth] << n
      end
      
      container = radix.flatten
    end

    return container
  end

  private
  
  def calckth(n, i)
    while(i > 1)
      n = n / 10
      i = i - 1
    end
    n % 10
  end

10、Bucket sort 桶排序

  # Bucket sort 桶排序
  # 将数组分到有限数量的桶里,每个桶再进行排序
  # Time Complexity: О(n) for best, О(n+k) for average, O(n^2) for worst-case
  # Space Complexity: О(n*k) for worst-case
  # Stable: Yes
  def bucketsort(container)
    bucket = []
    (0..9).each do |j|
      bucket[j] = []
    end
      
    container.each do |n|  
      k = getfirstnumber(n)
      bucket[k] << n
    end  
    
    (0..9).each do |j|  
      bucket[j] = quicksortB(bucket[j])  
    end
    
    bucket.flatten  
  end

  private
  
  #假设都是两位正整数
  def getfirstnumber(n)  
    m = n/10
    m=0 if m<0
    m=9 if m>9
    
    return m
  end
  
  def quicksortB(container)  
    (x=container.pop) ? quicksortB(container.select{|i| i <= x}) + [x] + quicksortB(container.select{|i| i > x}) : []  
  end

排序算法Ruby Part1

1、Bubble sort 冒泡排序

  # Bubble sort 冒泡排序
  # 每次扫描,比较和交换相邻元素,保证一次扫描后,最大元素在最后一个
  # 每次扫描后,扫描队列长度减一
  # Time Complexity: О(n^2)
  # Space Complexity: О(n) total, O(1) auxiliary
  # Stable: Yes
  def bubble_sort(container)
    swap=true
    while(swap)
      swap=false
      for i in 0..(container.size-1) do
        for j in 0..i do
          if(container[i]<container[j])
            r=container[j]
            container[j]=container[i]
            container[i]=r
            swap=true
          end
        end
      end
    end

    return container
  end

2、Selection sort 选择排序

  # Selection sort 选择排序
  # 从头扫描到尾,每次将最小元素放到第一个
  # 每次扫描后,队列长度减一
  # Time Complexity: О(n^2)
  # Space Complexity: О(n) total, O(1) auxiliary
  # Stable: Yes
  def selection_sort(container)
    for i in 0..container.size-1 do
      min = i
      for j in i..container.size-1 do
        if(container[min]>container[j])
          min=j
        end
      end

      r=container[i]
      container[i]=container[min]
      container[min]=r

    end

    return container

  end

3、Insertion sort 插入排序

  # Insertion sort 插入排序
  # 起始队列长度为1,每次向队列增加1个数字
  # 通过元素移动,将队列调整为有序状态
  # Time Complexity: О(n^2)
  # Space Complexity: О(n) total, O(1) auxiliary
  # Stable: Yes
  def insertion_sort(container)
    if container.size <2
      return container
    end

    for i in 1..container.size-1 do
      val = container[i]
      j=i-1
      while(j>=0 and container[j]>val) do
        container[j+1]=container[j]
        j=j-1
      end

      container[j+1] = val
    end

    return container
  end

4、Shell Sort 希尔排序

  # Shell Sort 希尔排序
  # 指定一个步长,将需要排序的数字,按序号%步长分组
  # 对每组进行插入排序,然后减小步长,再次分组,排序
  # 直到步长为1结束
  # Time Complexity: О(n^2)
  # Space Complexity: О(n) total, O(1) auxiliary
  # Stable: Yes
  def shell_sort(container)
    step = container.size/2
    while step>0 do
      for i in step..container.size-1 do
        val= container[i]
        j=i-step
        while(j>=0 and container[j]>val) do
          container[j+step] = container[j]
          j=j-step
        end
        container[j+step]=val
      end

      #puts(">>")
      #puts(container)

      step = step/2
    end

    return container
  end

5、Merge sort 归并排序

  # Merge sort 归并排序
  # 二路归并首先将归并长度定为2,在归并集内进行排序
  # 然后归并长度×2,将有序集合进行归并
  # Time Complexity: О(nlogn) average and worst-case
  # Space Complexity: О(n) auxiliary
  # Stable: Yes
  def mergesort(container)
    return container if container.size <= 1

    mid = container.size / 2
    left = container[0...mid]
    right = container[mid...container.size]

    merge(mergesort(left), mergesort(right))
  end

  private

  def merge(left, right)
    sorted = []
    until left.empty? or right.empty?
      left.first <= right.first ? sorted << left.shift : sorted << right.shift
    end
    sorted + left + right
  end