如何在Java中正确实现快速排序的递归？

我目前正在用Java进行快速排序，我已经成功地为第一次迭代排序了列表。尽管如此，我的递归实现并不是我想要的。原因是什么

列表是[11,4,53,65,44,23,202,37,1]

以下是按Java升序正确实现的快速排序

 public static List<Integer> quickSort(List<Integer> l1, int from, int to) {
        System.out.println("Quick Sort \n");

        long startTime = System.nanoTime();
        //select a pivot - the last element of the list
        int pivot = l1.get(to - 1);
        //introduce two counters:
        int counterLastSwapPos = 0;//this first one will track the index of the element
        //that is bigger than the pivot - we start from zero (we never actually
        // know that this number is actually bigger - it is a
        // presupposition)
        for (int counter = 0; counter < l1.indexOf(pivot); counter++) {
            //we also have a counter to track our position during the iteration
            //if the element at the current position is smaller than the pivot
            //swap the element(current position) with the element that is bigger
            //than the pivot.
            if (l1.get(counter) < pivot) {
                int temp = l1.get(counter);
                l1.set(counter, l1.get(counterLastSwapPos));
                l1.set(counterLastSwapPos, temp);
                //Once the swap has happened - increment the counter
                //that tracks the number bigger than the pivot
                counterLastSwapPos++;
                //finally, in the loop, the position counter will be
                //automatically incremented
            }
            //when the position counter reaches the last allowed position,
            //swap the pivot with the the counter that tracks
            // the number bigger than the pivot
            if (counter == l1.indexOf(pivot) - 1) {
                l1.set(l1.indexOf(pivot), l1.get(counterLastSwapPos));
                l1.set(counterLastSwapPos, pivot);
            }
        }
        //as this sorting is a "Divide&Conquer" type, we use recursion to perform
        //the same operations on two parts of the list. That is why, (if you scroll up),
        //you'll see that once the list becomes size of 1, the recursion will stop.
        //Our pivot is now somewhere in the middle - this was our aim.
        //Now, pay attention to perform the recursion on
        //two lists that WILL NOT include the pivot itself
        if (from < l1.indexOf(pivot)) quickSort(l1, from, l1.indexOf(pivot));
        if (l1.indexOf(pivot) + 1 < to) quickSort(l1, l1.indexOf(pivot) + 1, to);
        //list is sorted
        long endTime = System.nanoTime();
        long duration = (endTime - startTime);
        System.out.println("Time: " + duration + "\n");

        return l1;
    }

---

顺便说一下，你要么随意选择你的支点，要么把它放在中间和右边，而不是把它放在最后。to@user真正地关于这个解释视频，你能说些什么？@用户我已经试过使用子列表了，但是我如何将它们加在一起呢？我只想操纵原始列表，这样我就不必像合并排序那样考虑如何连接排序后的列表。例如，我不确定，但就我所见，该视频似乎没有使用与您相同的算法。看这个：你是什么意思？在这儿吗？quickSortl1，0，l1.indexOfpivot；quickSortl1，l1.indexOfpivot+1，l1.size；如果是，请执行此quickSortl1，from，l1.indexOfpivot；快速入门l1，l1.indexOfpivot+1，至；

 public static List<Integer> quickSort(List<Integer> l1, int from, int to) {
        System.out.println("Quick Sort \n");

        long startTime = System.nanoTime();
        //select a pivot - the last element of the list
        int pivot = l1.get(to - 1);
        //introduce two counters:
        int counterLastSwapPos = 0;//this first one will track the index of the element
        //that is bigger than the pivot - we start from zero (we never actually
        // know that this number is actually bigger - it is a
        // presupposition)
        for (int counter = 0; counter < l1.indexOf(pivot); counter++) {
            //we also have a counter to track our position during the iteration
            //if the element at the current position is smaller than the pivot
            //swap the element(current position) with the element that is bigger
            //than the pivot.
            if (l1.get(counter) < pivot) {
                int temp = l1.get(counter);
                l1.set(counter, l1.get(counterLastSwapPos));
                l1.set(counterLastSwapPos, temp);
                //Once the swap has happened - increment the counter
                //that tracks the number bigger than the pivot
                counterLastSwapPos++;
                //finally, in the loop, the position counter will be
                //automatically incremented
            }
            //when the position counter reaches the last allowed position,
            //swap the pivot with the the counter that tracks
            // the number bigger than the pivot
            if (counter == l1.indexOf(pivot) - 1) {
                l1.set(l1.indexOf(pivot), l1.get(counterLastSwapPos));
                l1.set(counterLastSwapPos, pivot);
            }
        }
        //as this sorting is a "Divide&Conquer" type, we use recursion to perform
        //the same operations on two parts of the list. That is why, (if you scroll up),
        //you'll see that once the list becomes size of 1, the recursion will stop.
        //Our pivot is now somewhere in the middle - this was our aim.
        //Now, pay attention to perform the recursion on
        //two lists that WILL NOT include the pivot itself
        if (from < l1.indexOf(pivot)) quickSort(l1, from, l1.indexOf(pivot));
        if (l1.indexOf(pivot) + 1 < to) quickSort(l1, l1.indexOf(pivot) + 1, to);
        //list is sorted
        long endTime = System.nanoTime();
        long duration = (endTime - startTime);
        System.out.println("Time: " + duration + "\n");

        return l1;
    }