Tag Archives: Algorithm

排序算法C Part01

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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

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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

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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

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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

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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

二进制数字中1的个数

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#include "stdafx.h"
#include <nmmintrin.h>

/*
 *通过移位计算1的个数
 *每一位都要判断和处理
 */
int BitCount(unsigned int n)
{
	unsigned int c = 0;
	while (n >0)
	{
		if ((n & 1) == 1)++c;
		n >>= 1;
	}
	return c;
}

/*
 *通过减法及位运算，保证每次至少消除一个1
 *只处理1的位，0的位不处理
 */
int BitCountWithMinus(unsigned int n)
{
	unsigned int c = 0;
	for (c = 0; n; ++c)
	{
		n &= (n - 1);
	}
	return c;
}

/*
*将32位数字，截为4个8位数
*通过查表，得到每个8位数中1的个数，然后求和
*/
int BitCountLUT8(unsigned int n)
{
	unsigned char BitsSetTable256[256] = { 0 };

	for (int i = 0; i <256; i++)
	{
		BitsSetTable256[i] = (i & 1) + BitsSetTable256[i / 2];
	}

	unsigned int c = 0;

	unsigned char* p = (unsigned char*)&n;

	c = BitsSetTable256[p[0]] +
		BitsSetTable256[p[1]] +
		BitsSetTable256[p[2]] +
		BitsSetTable256[p[3]];

	return c;
}

/*
*将32位数字，截为4个8位数
*通过查表，得到每个8位数中1的个数，然后求和
*LUT表已经计算好
*/
int BitCountLUT8Static(unsigned int n)
{
	unsigned int table[256] =
	{
		0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
		1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
		1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
		2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
		1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
		2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
		2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
		3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
		1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
		2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
		2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
		3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
		2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
		3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
		3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
		4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8,
	};

	return table[n & 0xff] +
		table[(n >> 8) & 0xff] +
		table[(n >> 16) & 0xff] +
		table[(n >> 24) & 0xff];
}

/*
*将32位数字，截为8个4位数
*通过查表，得到每个4位数中1的个数，然后求和
*LUT表已经计算好
*/
int BitCountLUT4Static(unsigned int n)
{
	unsigned int table[16] =
	{
		0, 1, 1, 2,
		1, 2, 2, 3,
		1, 2, 2, 3,
		2, 3, 3, 4
	};

	unsigned int count = 0;
	while (n)
	{
		count += table[n & 0xf];
		n >>= 4;
	}
	return count;
}

/*
*将32位数字，相邻的2位求和，然后相邻的4位求和
*然后相邻的8位、16位、32位求和
*/
int BitCountParallel(unsigned int n)
{
	n = (n & 0x55555555) + ((n >> 1) & 0x55555555);
	n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
	n = (n & 0x0f0f0f0f) + ((n >> 4) & 0x0f0f0f0f);
	n = (n & 0x00ff00ff) + ((n >> 8) & 0x00ff00ff);
	n = (n & 0x0000ffff) + ((n >> 16) & 0x0000ffff);

	return n;
}

/*
*第一行，计算每三位的1的个数（其实11*3是33位，但最高一位可以假设为0，所以没问题）
*第二行，实际上是先计算了相邻6位中1的个数（不会产生进位，结果最多占到3位），并将前三位置为0
*然后通过取模，相当于将每6位数字做了加法
*/
int BitCountMagic(unsigned int n)
{
	unsigned int tmp = n - ((n >> 1) & 033333333333) - ((n >> 2) & 011111111111);
	return ((tmp + (tmp >> 3)) & 030707070707) % 63;
}

unsigned int n = 127;
unsigned int bitCount = _mm_popcnt_u32(n);


/*
*将8位数字，转换为MY_UNSIGHED_CHAR结构体，然后求和
*/
struct MY_UNSIGHED_CHAR
{
	unsigned a : 1;
	unsigned b : 1;
	unsigned c : 1;
	unsigned d : 1;
	unsigned e : 1;
	unsigned f : 1;
	unsigned g : 1;
	unsigned h : 1;
};

long BitCountStuct(unsigned char b)
{
	struct MY_UNSIGHED_CHAR *by = (struct MY_UNSIGHED_CHAR*)&b;
	return (by->a + by->b + by->c + by->d + by->e + by->f + by->g + by->h);
}

int _tmain(int argc, _TCHAR* argv[])
{
	printf("There are %d 1 in 133\n", BitCount(133));
	printf("There are %d 1 in 133\n", BitCountWithMinus(133));
	printf("There are %d 1 in 133\n", BitCountLUT8(133));
	printf("There are %d 1 in 133\n", BitCountLUT8Static(133));
	printf("There are %d 1 in 133\n", BitCountLUT4Static(133));
	printf("There are %d 1 in 133\n", BitCountParallel(133));
	printf("There are %d 1 in 133\n", BitCountMagic(133));
	printf("There are %d 1 in 133\n", BitCountStuct(133));
	return 0;
}

用位运算实现加减乘除

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        //加法
	public static int add(int a, int b)  
	{  
	    int ans;  
	    while(b!=0)  
	    {
	        ans = a^b;
	        b = ((a&b)<<1);
	        a = ans;  
	    }  
	    return a;  
	}  
	
	//减法
	public static int sub(int a, int b)
	{  
	    return add(a, -b);  
	}
	
	//正数乘法
	private static int posMultiply(int a,int b)  
	{  
	    int ans = 0;  
	    while(b>0)  
	    {  
	        if((b&0x1)==1)ans = add(ans, a);  
	        a = (a<<1);  
	        b = (b>>1);  
	    }  
	    return ans;  
	}  
	  
	//乘法
	public static int multiply(int a,int b)  
	{  
	    if(a==0||b==0)return 0;  
	    if(a>0 && b>0)  
	        return posMultiply(a, b);  
	    if(a<0)  
	    {  
	        if(b<0)  
	        {  
	            return posMultiply(-a, -b);  
	        } 
	        else
	        {
	        	return -posMultiply(-a, b ); 
	        }
	    }
	    else
	    {
	    	return -posMultiply(a, -b);
	    }
	}  
	  
	//正数除法
	private static int posDiv(int x,int y)  
	{  
	    int ans=0;  
	    for(int i=31;i>=0;i--)  
	    {
	        if((x>>i)>=y)  
	        {  
	            ans+=(1<<i);  
	            x-=(y<<i);  
	        }  
	    }  
	    return ans;  
	}  
	  
	//除法
	public static int div( int a, int b )  
	{
		assert(b!=0);
	    if(a==0)return 0;

	    if(a>0)  
	    {  
	        if(b>0)
	        {
	            return posDiv(a,b); 
	        }
	        else
	        {
	        	return -posDiv( a,-b);  
	        }
	    }  
	    else
	    {
	    	if(b>0)
	    	{
		        return -posDiv(a, b);  
	    	}
	    	else
	    	{
	    		return -posDiv(-a, -b);  
	    	}
	    }
	    
	}   
	
	//求负数
	public static int negtive(int a)
	{  
	    return add(~a, 1);  
	}
	  
	//比较正数大小
	private static boolean isbigerPos( int a, int b )   
	{
	    int c = 1;  
	    b = (a^b);  
	    if(b==0)return false;
	    
	    while(b>0)  
	    {  
	    	b>>=1;
	        c <<= 1;  
	    }  
	    return (c&a)==0;  
	}   
	  
	//比较大小   
	public static boolean isbiger( int a, int b )   
	{
	    if(a<0)  
	    {  
	        if(b<0)  
	        {  
	            return isbigerPos( negtive(b), negtive(a) );  
	        }
	        else
	        {
	        	return false;  
	        }
	    }
	    else
	    {
		    if(b<0)
		    {
		        return true;  
		    }
		    else
		    {
		    	return isbigerPos(a, b);  
		    }
	    }

	}

	public static int divideby3(int x)  
	{  
	    int sum = 0;  
	    while(x > 3)  
	    {  
	        sum = add(x>>2 , sum);  
	        x = add(x>>2 , x&3);  
	    } 
	    if(x == 3) 
	    {
	        sum = add(sum , 1);
	    }
	    return sum;  
	}  
	
	public static void main(String[] args)
	{
		System.out.println(add(13,19));
		System.out.println(sub(13,19));
		System.out.println(multiply(13,19));
		System.out.println(div(10,5));
		System.out.println(negtive(13));
		System.out.println(isbiger(10,5));
		System.out.println(isbiger(10,11));
		
		System.out.println(divideby3(15));
		System.out.println(divideby3(14));
		System.out.println(divideby3(13));
		System.out.println(divideby3(12));
		System.out.println(divideby3(11));
	}

几种常见求解平方根的方法

Posted on by — No Comments ↓ 
        //精度
	private final static double accuracy= 1e-6;

	/**
	 * 暴力求解
	 */
	public static double bruteSqrt(double x)
	{
		assert(x>=0);
		double ans=0.0;
		while (Math.abs(x - ans * ans) > accuracy)ans += accuracy;
		return ans;
	}

	/**
	 * 牛顿法求解
	 */
	public static double newtonSqrt(double x)
	{
		assert(x>=0);
		double avg = x;
		double last_avg = Double.MAX_VALUE;

		while (Math.abs(avg - last_avg) > accuracy)
		{
			last_avg = avg;
			avg = (avg + x / avg) / 2;
		}
		return avg;
	}

	/**
	 * 二分法求解
	 */
	public static double binarySqrt(double x)
	{
		assert(x>=0);

		double low = 0;
		double high = x;
		double mid = Double.MAX_VALUE;
		double last_mid = Double.MIN_VALUE;

		while (Math.abs(mid - last_mid) > accuracy)
		{
			last_mid = mid;
			mid = (low + high)/2;
			if (mid*mid>x)high = mid;
			if (mid*mid<x)low = mid;
		}
		return mid;

	}

	private final static int[] LUT =
	{ 0, 16, 22, 27, 32, 35, 39, 42, 45, 48, 50, 53, 55, 57, 59, 61, 64, 65, 67, 69, 71, 73, 75, 76, 78, 80, 81, 83, 84,
			86, 87, 89, 90, 91, 93, 94, 96, 97, 98, 99, 101, 102, 103, 104, 106, 107, 108, 109, 110, 112, 113, 114, 115,
			116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 128, 128, 129, 130, 131, 132, 133, 134, 135, 136,
			137, 138, 139, 140, 141, 142, 143, 144, 144, 145, 146, 147, 148, 149, 150, 150, 151, 152, 153, 154, 155,
			155, 156, 157, 158, 159, 160, 160, 161, 162, 163, 163, 164, 165, 166, 167, 167, 168, 169, 170, 170, 171,
			172, 173, 173, 174, 175, 176, 176, 177, 178, 178, 179, 180, 181, 181, 182, 183, 183, 184, 185, 185, 186,
			187, 187, 188, 189, 189, 190, 191, 192, 192, 193, 193, 194, 195, 195, 196, 197, 197, 198, 199, 199, 200,
			201, 201, 202, 203, 203, 204, 204, 205, 206, 206, 207, 208, 208, 209, 209, 210, 211, 211, 212, 212, 213,
			214, 214, 215, 215, 216, 217, 217, 218, 218, 219, 219, 220, 221, 221, 222, 222, 223, 224, 224, 225, 225,
			226, 226, 227, 227, 228, 229, 229, 230, 230, 231, 231, 232, 232, 233, 234, 234, 235, 235, 236, 236, 237,
			237, 238, 238, 239, 240, 240, 241, 241, 242, 242, 243, 243, 244, 244, 245, 245, 246, 246, 247, 247, 248,
			248, 249, 249, 250, 250, 251, 251, 252, 252, 253, 253, 254, 254, 255 };

	/**
	 * 查表法求解
	 */
	public static int intLutSqrt(int x)
	{
		int xn;

		if (x >= 0x10000)
		{
			if (x >= 0x1000000)
			{
				if (x >= 0x10000000)
				{
					if (x >= 0x40000000)
					{
						xn = LUT[x >> 24] << 8;
					}
					else
					{
						xn = LUT[x >> 22] << 7;
					}
				}
				else
				{
					if (x >= 0x4000000)
					{
						xn = LUT[x >> 20] << 6;
					}
					else
					{
						xn = LUT[x >> 18] << 5;
					}
				}

				xn = (xn + 1 + (x / xn)) >> 1;
				xn = (xn + 1 + (x / xn)) >> 1;
				return ((xn * xn) > x) ? --xn : xn;
			}
			else
			{
				if (x >= 0x100000)
				{
					if (x >= 0x400000)
					{
						xn = LUT[x >> 16] << 4;
					}
					else
					{
						xn = LUT[x >> 14] << 3;
					}
				}
				else
				{
					if (x >= 0x40000)
					{
						xn = LUT[x >> 12] << 2;
					}
					else
					{
						xn = LUT[x >> 10] << 1;
					}
				}

				xn = (xn + 1 + (x / xn)) >> 1;

				return ((xn * xn) > x) ? --xn : xn;
			}
		}
		else
		{
			if (x >= 0x100)
			{
				if (x >= 0x1000)
				{
					if (x >= 0x4000)
					{
						xn = (LUT[x >> 8]) + 1;
					}
					else
					{
						xn = (LUT[x >> 6] >> 1) + 1;
					}
				}
				else
				{
					if (x >= 0x400)
					{
						xn = (LUT[x >> 4] >> 2) + 1;
					}
					else
					{
						xn = (LUT[x >> 2] >> 3) + 1;
					}
				}

				return ((xn * xn) > x) ? --xn : xn;
			}
			else
			{
				if (x >= 0)
				{
					return LUT[x] >> 4;
				}
			}
		}

		return -1;
	}

	/**
	 * Quake III中快速求解平方根倒数的方法
	 */
	public static float fastInvSqrt(float x)  
	{  
	     float xhalf = 0.5f*x;  
	     int f2i = Float.floatToRawIntBits(x);
	     f2i = 0x5f375a86-(f2i>>1);
	     x = Float.intBitsToFloat(f2i);
	     x = x*(1.5f-xhalf*x*x);
	     x = x*(1.5f-xhalf*x*x);
	     return x;  
	}
	
	/**
	 * Quake III中快速求解平方根方法
	 */
	public static float fastSqrt(float x) {  
	    float y=x;
	    float xhalf = 0.5f*x;
	    int f2i = Float.floatToRawIntBits(x);  
	    f2i = 0x5f3759df-(f2i>>1);  
	    x = Float.intBitsToFloat(f2i);  
	    x  = x * (1.5f-(xhalf*x*x));  
	    x  = x * (1.5f-(xhalf*x*x));  
	    return y*x;  
	}

	public static void main(String[] args)
	{
		System.out.println(bruteSqrt(3));
		System.out.println(newtonSqrt(3));
		System.out.println(binarySqrt(3));
		System.out.println(intLutSqrt(64));
		System.out.println(1/fastInvSqrt(3));
		System.out.println(fastSqrt(3));
	}