Java中的Inplace Quicksort [英] Inplace Quicksort in Java

问题描述

为了刷新一些Java，我试图实现一个可以对整数数组进行排序的快速排序（inplace）算法。以下是我到目前为止的代码。您可以通过 sort（a，0，a.length-1）调用它。

For refreshing some Java I tried to implement a quicksort (inplace) algorithm that can sort integer arrays. Following is the code I've got so far. You can call it by sort(a,0,a.length-1).

此代码显然失败（进入无限循环）如果两个'指针' i，j 将每个指向与数据透视表具有相同值的数组条目。 pivot元素 v 始终是当前分区中最右边的分区（索引最大的分区）。

This code obviously fails (gets into an infinite loop) if both 'pointers' i,j point each to an array entry that have the same values as the pivot. The pivot element v is always the right most of the current partition (the one with the greatest index).

但我无法弄清楚如何避免这种情况，是否有人看到了解决方案？

But I just cannot figure out how to avoid that, does anyone see a solution?

static void sort(int a[], int left, int right)   {
    if (right > left){
        int i=left, j=right-1, tmp;
        int v = a[right]; //pivot
        int counter = 0;
        do {
            while(a[i]<v)i++;
            while(j>0 && a[j]>v)j--;

            if( i < j){
                tmp = a[i];
                a[i] = a[j];
                a[j] = tmp;
            }
        } while(i < j);
        tmp = a[right];
        a[right] = a[i];
        a[i] = tmp;
        sort(a,left,i-1);
        sort(a,i+1,right);

    }
}    

推荐答案

这应该有用（将检查一点的正确性，它可以工作！）：

This should work (will check for correctness in a bit, it works!):

编辑：我以前在错误检查中犯了一个错误。我忘了添加2个条件，这里是修改后的代码。

I previously made a mistake in error checking. I forgot to add 2 more conditions, here is the amended code.

public static void main (String[] args) throws java.lang.Exception
{
    int b[] = {10, 9, 8, 7, 7, 7, 7, 3, 2, 1};
    sort(b,0,b.length-1);
    System.out.println(Arrays.toString(b));
}

static void sort(int a[], int left, int right)   {  
   if (right > left){
    int i=left, j=right, tmp;    
    //we want j to be right, not right-1 since that leaves out a number during recursion

    int v = a[right]; //pivot

    do {
        while(a[i]<v)
          i++;
        while(a[j]>v) 
        //no need to check for 0, the right condition for recursion is the 2 if statements below.
          j--;

        if( i <= j){            //your code was i<j
           tmp = a[i];
           a[i] = a[j];
           a[j] = tmp;
           i++;            
           j--;
           //we need to +/- both i,j, else it will stick at 0 or be same number
        }
   } while(i <= j);           //your code was i<j, hence infinite loop on 0 case

    //you had a swap here, I don't think it's needed.
    //this is the 2 conditions we need to avoid infinite loops
    // check if left < j, if it isn't, it's already sorted. Done

    if(left < j)  sort(a,left,j);
    //check if i is less than right, if it isn't it's already sorted. Done
    // here i is now the 'middle index', the slice for divide and conquer.

    if(i < right) sort(a,i,right);
  }

}

IDEOne在线编译器中的此代码

基本上我们确保如果i / j的值与pivot相同，我们也交换值，并且突破递归。

Basically we make sure that we also swap the value if the value of i/j is the same as the pivot, and break out of the recursion.

此外还检查了伪代码的长度，好像我们有一个只有1个项目的数组已经排序了（我们忘记了基本情况），我认为我们需要它，但是因为你传入了索引和整个数组，而不是子数组，我们只增加i和j所以算法不会坚持0（它们已经完成排序）但仍然继续排序1的数组。:)

Also there was a check in the pseudocode for the length, as if we have an array of just 1 item it's already sorted (we forgot the base case), I thought we needed that but since you pass in the indexes and the entire array, not the subarray, we just increment i and j so the algorithm won't stick at 0 (they're done sorting) but still keep sorting an array of 1. :)

另外，我们必须添加2个条件来检查数组是否已经为递归调用排序。没有它，我们将永远排序已排序的数组，因此另一个无限循环。看看我是如何添加检查，如果留下小于j，如果我小于正确。此外，在传递i和j的那一点上，我实际上是我们为分而治之分裂的中间指数，而j将是中间值之前的值。

Also, we had to add 2 conditions to check if the array is already sorted for the recursive calls. without it, we'll end up sorting an already sorted array forever, hence another infinite loop. see how I added checks for if left less than j and if i less than right. Also, at that point of passing in i and j, i is effectively the middle index we split for divide and conquer, and j would be the value right before the middle value.

它的伪代码取自 RosettaCode ：

function quicksort(array)
    if length(array) > 1
        pivot := select any element of array
        left := first index of array
        right := last index of array
        while left ≤ right
            while array[left] < pivot
                left := left + 1
            while array[right] > pivot
                right := right - 1
            if left ≤ right
                swap array[left] with array[right]
                left := left + 1
                right := right - 1
        quicksort(array from first index to right)
        quicksort(array from left to last index)

参考：此 SO问题

同时阅读这是一个快速的复习，它与oridnary while循环实现不同

这很有趣:)

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