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Stackoverflow 与 Quicksort Java 实现

[英]Stackoverflow with Quicksort Java implementation

Having some problems implementing quicksort in java.在 Java 中实现快速排序时遇到一些问题。 I get a stackoverflow error when I run this program and I'm not exactly sure why.运行此程序时出现计算器溢出错误,但我不确定原因。 If anyone can point out the error, it would be great.如果有人能指出错误,那就太好了。

si is the starting index. si 是起始索引。 ei is the ending index. ei 是结束索引。

public static void qsort(int[] a, int si, int ei){

    //base case
    if(ei<=si || si>=ei){}


    else{ 
        int pivot = a[si]; 
        int length = ei - si + 1; 
        int i = si+1; int tmp; 

        //partition array 
        for(int j = si+1; j<length; j++){
            if(pivot > a[j]){
                tmp = a[j]; 
                a[j] = a[i]; 
                a[i] = tmp; 

                i++; 
            }
        }

        //put pivot in right position
        a[si] = a[i-1]; 
        a[i-1] = pivot; 

        //call qsort on right and left sides of pivot
        qsort(a, 0, i-2); 
        qsort(a, i, a.length-1); 
    }
}

First you should fix the bounds of the qsort recursive call as suggested by Keith, since otherwise you're always sorting the whole array over and over again.首先,您应该按照 Keith 的建议修复 qsort 递归调用的边界,否则您总是一遍又一遍地对整个数组进行排序。 The you must adjust your partition loop: j is an index, going from the beginning of the subarray to the end of it (including the last element).你必须调整你的分区循环:j 是一个索引,从子数组的开头到它的结尾(包括最后一个元素)。 So you must loop from si + 1 to ei (including ei).所以你必须从 si + 1 循环到 ei(包括 ei)。

So this is the corrected code.所以这是更正后的代码。 I ran a few test cases and it seems to sort just fine.我运行了一些测试用例,它似乎排序得很好。

    public static void qsort(int[] a, int si, int ei){
    //base case
    if(ei<=si || si>=ei){}

    else{ 
        int pivot = a[si]; 
        int i = si+1; int tmp; 

        //partition array 
        for(int j = si+1; j<= ei; j++){
            if(pivot > a[j]){
                tmp = a[j]; 
                a[j] = a[i]; 
                a[i] = tmp; 

                i++; 
            }
        }

        //put pivot in right position
        a[si] = a[i-1]; 
        a[i-1] = pivot; 

        //call qsort on right and left sides of pivot
        qsort(a, si, i-2); 
        qsort(a, i, ei); 
    }
}
int partition(int array[], int too_big_index, int too_small_index)
{
     int x = array[too_big_index];
     int i = too_big_index;
     int j = too_small_index;
     int temp;
     do
     {                 
         while (x <array[j])
        {
              j --;
        } 
         while (x >array[i])
         {
              i++;
         } 
          if (i < j)
         { 
                 temp = array[i];    
                 array[i] = array[j];
                 array[j] = temp;
         }
     }while (i < j);     
     return j;           // middle  
}

void QuickSort(int num[], int too_big_index, int too_small_index)
{
      // too_big_index =  beginning of array
      // too_small_index = end of array

     int middle;
     if (too_big_index < too_small_index)
    {
          middle = partition(num, too_big_index, too_small_index);
          QuickSort(num, too_big_index, middle);   // sort first section
          QuickSort(num, middle+1, too_small_index);    // sort second section
     }
     return;
}



void main()
{
    int arr[]={8,7,13,2,5,19,1,40,12,34};

    QuickSort(arr,0,9);
    for(int i=0;i<10;i++)
         System.out.println(arr[i]);
}

You might have an unbounded recursion bug on your hands.您可能手头上有一个无限递归错误。 Not sure from my quick scan of it, but...从我的快速扫描中不确定,但是......

Even if you don't, you're still going to use lots of stack with this implementation.即使您不这样做,您仍然会在此实现中使用大量堆栈。 Enough to cause a stack overflow.足以导致堆栈溢出。 What happens if you call it with 1 million items that are already sorted?如果你用 100 万个已经排序的项目调用它会发生什么? You'll partition them into 1 and 999,999 items, then recurse.您将它们划分为 1 和 999,999 个项目,然后递归。 So you'll need 1 million stack frames to make this work.因此,您需要 100 万个堆栈帧才能完成这项工作。

There are lots of ways to solve this, including recursing on the smaller of the two ranges and iterating on the larger of the two, or implementing the stack yourself in a heap datastructure, etc. You probably want to do even better than that, though, as the deep stack also means you're blowing past the O(n lg n) sorting bound.有很多方法可以解决这个问题,包括在两个范围中的较小者上进行递归并在两个范围中的较大者上进行迭代,或者自己在堆数据结构中实现堆栈等。不过,您可能想要做得比这更好,因为深堆栈也意味着你正在吹过 O(n lg n) 排序界限。

ps the bug is here: ps错误在这里:

qsort(a, 0, i-2); 
qsort(a, i, a.length-1); 

should be应该

qsort(a, si, i-2);
qsort(a, i, ei);

You can try this:你可以试试这个:

public void sort(int[] A) {
        if (A == null || A.length == 0)
            return;
        quicksort(A, 0, A.length - 1);
    }

    public void quicksort(int[] A, int left, int right) {
        int pivot = A[left + (right - left) / 2];
        int i = left;
        int j = right;
        while (i <= j) {
            while (A[i] < pivot) {
                i++;
            }
            while (A[j] > pivot) {
                j--;
            }
            if (i <= j) {
                exchange(i, j);
                i++;
                j--;
            }
        }

        if(left < j)
            quicksort(A,left,j);
        if(i < right)
            quicksort(A,i,right);
    }

    public void exchange(int i, int j){
        int temp=A[i];
        A[i]=A[j];
        A[j]=temp;
    }

    public String toString() {
        String s = "";
        s += "[" + A[0];
        for (int i = 1; i < A.length; i++) {
            s += ", " + A[i];
        }
        s += "]";
        return s;
    }

Source: Code 2 Learn: Quick Sort Algorithm Tutorial来源: Code 2 学习:快速排序算法教程

import java.util.Arrays;


public class QuickSort {


    public static int pivot(int[] a, int lo, int hi){
        int mid = (lo+hi)/2;
        int pivot = a[lo] + a[hi] + a[mid] - Math.min(Math.min(a[lo], a[hi]), a[mid]) - Math.max(Math.max(a[lo], a[hi]), a[mid]);

        if(pivot == a[lo])
            return lo;
        else if(pivot == a[hi])
            return hi;
        return mid;
    }

    public static int partition(int[] a, int lo, int hi){

        int k = pivot(a, lo, hi);
        //System.out.println(k);
        swap(a, lo, k);
        //System.out.println(a);
        int j = hi + 1;
        int i = lo;
        while(true){

            while(a[lo] < a[--j])
                if(j==lo)   break;

            while(a[++i] < a[lo])
                if(i==hi) break;

            if(i >= j)  break;
            swap(a, i, j);
        }
        swap(a, lo, j);
        return j;
    }

    public static void sort(int[] a, int lo, int hi){
        if(hi<=lo)  return;
        int p = partition(a, lo, hi);
        sort(a, lo, p-1);
        sort(a, p+1, hi);
    }

    public static void swap(int[] a, int b, int c){
        int swap = a[b];
        a[b] = a[c];
        a[c] = swap;
    }

    public static void sort(int[] a){
        sort(a, 0, a.length - 1);
        System.out.print(Arrays.toString(a));
    }

    public static void main(String[] args) {
        int[] arr = {5,8,6,4,2,9,7,5,9,4,7,6,2,8,7,5,6};
        sort(arr);
    }
}

Try this.试试这个。 It will work for sure.它肯定会起作用。

//Just implemented the tester class for this and it'll work //刚刚为此实现了测试器类,它会起作用

public int[] sort(int[] A, int from, int to ){ public int[] sort(int[] A, int from, int to ){

if(from<to){
    int pivot=partition(A,from,to);
    if(pivot>1)
        sort(A,from, pivot-1);

    if(pivot+1<to)
        sort(A, pivot+1, to);


}

return array;

} }

public int partition(int A[ ], int from, int to){ public int partition(int A[], int from, int to){

while(from < to){
    int pivot=A[from];

    while(A[from]<pivot)
        from++;

    while(A[to]>pivot)
        to--;


    if(from<to)   
        swap(A,to,from);



}
    return to;
}

private void swap(int A[], int i, int j){
    int temp = A[i];
    A[i] = A[j];
    A[j] = temp;}

This should be quite spot on.这应该很到位。 Code below with this Image makes way more sense.下面带有此图像的代码更有意义。

public void quickSort(long[] a){

  int startingIndex = 0;
  int endingIndex = a.length - 1;
  qsort(a, startingIndex, endingIndex);

}

private void qsort(long[] a, int startingIndex, int endingIndex){

  if (startingIndex < endingIndex){
    int middleIndex = partition(a, startingIndex, endingIndex);
    qsort(a, startingIndex, middleIndex-1);
    qsort(a, middleIndex+1, endingIndex);
  }

}

private int partition(long[] a, int startingIndex, int endingIndex){
  long pivot = a[endingIndex];
  int endWall = endingIndex;
  int wall = 0;

  while (wall < endWall){

    if (a[wall] < pivot){
      wall++;
    }
    else {
      a[endWall] = a[wall];
      a[wall] = a[endWall - 1];
      a[endWall - 1] = pivot;
      endWall--;
    }
  }
  return wall;
}

Enjoy!享受!

public class MyQuickSort {

    private int array[];
    private int length;

    public void sort(int[] inputArr) {

        if (inputArr == null || inputArr.length == 0) {
            return;
        }
        this.array = inputArr;
        length = inputArr.length;
        quickSort(0, length - 1);
    }

    private void quickSort(int lowerIndex, int higherIndex) {

        int i = lowerIndex;
        int j = higherIndex;
        // calculate pivot number, I am taking pivot as middle index number
        int pivot = array[lowerIndex+(higherIndex-lowerIndex)/2];
        // Divide into two arrays
        while (i <= j) {
            /**
             * In each iteration, we will identify a number from left side which 
             * is greater then the pivot value, and also we will identify a number 
             * from right side which is less then the pivot value. Once the search 
             * is done, then we exchange both numbers.
             */
            while (array[i] < pivot) {
                i++;
            }
            while (array[j] > pivot) {
                j--;
            }
            if (i <= j) {
                exchangeNumbers(i, j);
                //move index to next position on both sides
                i++;
                j--;
            }
        }
        // call quickSort() method recursively
        if (lowerIndex < j)
            quickSort(lowerIndex, j);
        if (i < higherIndex)
            quickSort(i, higherIndex);
    }

    private void exchangeNumbers(int i, int j) {
        int temp = array[i];
        array[i] = array[j];
        array[j] = temp;
    }

    public static void main(String a[]){

        MyQuickSort sorter = new MyQuickSort();
        int[] input = {24,2,45,20,56,75,2,56,99,53,12};
        sorter.sort(input);
        for(int i:input){
            System.out.print(i);
            System.out.print(" ");
        }
    }
}
// the partition function
public static int Sort(int arr[], int start, int end)
{
    int pivot = arr[end];
    int pIndex = start;
    for(int i = start; i< end;i++)
    {
        if(arr[i] <= pivot)
        {
            int temp = arr[i];
            arr[i] = arr[pIndex];
            arr[pIndex] = temp;
            pIndex++;
        }
    }
    int temp = arr[pIndex];
    arr[pIndex] = pivot;
    arr[end] = temp;
    return pIndex;
}
public static void quickSort(int arr[],int start,int end)
{
    if(start>=end)
        return;
    // finding the pivot element
    int pivot = Sort(arr,start,end);
    quickSort(arr,start,pivot-1);
    quickSort(arr,pivot+1,end);
    
}

Quicksort is slightly sensitive to input that happens to be in the right order, in which case it can skip some swaps.快速排序对碰巧按正确顺序的输入稍微敏感,在这种情况下,它可以跳过一些交换。 Mergesort doesn't have any such optimizations, which also makes Quicksort a bit faster compared to Mergesort. Mergesort 没有任何此类优化,这也使 Quicksort 比 Mergesort 快一些。

Why Quick sort is better than Merge sort 为什么快速排序比合并排序好

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