Java脚本和 Java 中的快速排序

Quicksort in JavaScript and Java

刚刚发现一个奇怪的问题。我尝试使用Java脚本实现基于Java中编写的代码的快速排序,然而,结果不是我所期望的。 Java代码:

public class Sort {
    public static void main (String[] args){
        int[] number = {1,10,2,2,2,3,5,7,8,2};
        for (int i = 0; i < number.length ; i++) {
            System.out.println(quickSort(number)[i]);//1,2,2,2,2,3,5,7,8,10
        }
    }
    static int[] quickSort(int[] number){
        int length = number.length;
        int start = 0;
        int end = length -1;
        number = quick_sort (start,end,number);
        return number;
    }
    private static int[] quick_sort(int start,int end,int[] number) {

        int pivot = number[(start+end)/2];
        int s = start;
        int e = end;
        while (s <= e) {
            while(number[s] < pivot) {
                s++;
            }
            while(number[e] > pivot) {
                e--;
            }
            //swap the value
            if (s <= e) {
                int temp = number[s];
                number[s] = number[e];
                number[e] = temp;
                //move both sides cursor
                s++;
                e--;
            }

        }

        // recursively sorting lower half
        if (start < e) {
            quick_sort(start, e, number);
        }
        //recursively sorting higher half
        if (end > s) {
            quick_sort(s, end, number);
        }
        return number;
    }
}

Java脚本:

function main() {
    var numbers  = [1,10,2,2,2,3,5,7,8,2];
    var length = numbers.length;
    var start = 0;
    var end = length - 1;
    alert(quickSort(start, end, numbers));//2,3,5,8,7,2,1,2,2,10  <----- not sorted
}

function quickSort(start, end, numbers) {
    var s,e,temp,pivot;
    s = start;
    e = end;
    pivot = numbers[(start+end)/2];
    while(s<=e){
        while(numbers[s] < pivot ) {
            s++;
        }
        while(numbers[e] > pivot) {
            e--;
        }
        if(s<=e) {
            temp = numbers[s];
            numbers[s] = numbers[e];
            numbers[e] = temp;
            e--;
            s++;
        }
    }
    if(start < e) {
        quickSort(start, e, numbers);
    }
    if(end > s) {
        quickSort(s, end, numbers);
    }
    return numbers;
}
main();

写在Java中的代码给出了1,2,2,2,2,3,5,7,8,10的结果,在另一个hard中,写在Java中的代码脚本生成 2,3,5,8,7,2,1,2,2,10 的结果,这与原始数组不同,但仍未正确排序。对此有什么想法吗?提前致谢

编辑:

感谢您的回复,我正在研究逻辑而不是基础知识。

(start+end)/2 在 JS 中不一定是整数。它适用于 Math.floor((start+end)/2).