快速排序错误Java

Question

public int partition(int[]a,int i,int j){
    int x=a[i];
    int c=i;
    for (int d = c+1; d < j; d++) {
        if (a[d]<=x) {
            c=c+1;
            exchange(a, c, d);
        }

    }
    exchange(a, c, i);
    return c;
}
public void exchange(int[]a,int c,int d){
    int temp=a[c];
    a[c]=a[d];
    a[d]=temp;
}
public void sort(int[]a,int i,int j){
    int index;
    if(i<j){
        index=partition(a, i, j);
        sort(a, i, index-1);
        sort(a,index+1, j);
    }
}

**strong text**public static void main(String[] args) {
    // TODO code application logic here
    QUICKSORT abir=new QUICKSORT();
    abir.PRINT(abir.a);
    abir.sort(abir.a, 0,10);
    abir.PRINT(abir.a);
}

Here is my QuickSort code. If i enter [6 10 13 5 8 3 2 11] it prints [2 5 3 6 8 11 10 13]

Can anyone explain whats wrong with my code??? THank you

Answer 1

Your code seems quite strange to me, where did you take that? Anyway why don't you take a look at this page for an explanation of quick sort? http://www.algolist.net/Algorithms/Sorting/Quicksort

int partition(int arr[], int left, int right)
{
      int i = left, j = right;
      int tmp;
      int pivot = arr[(left + right) / 2];

      while (i <= j) {
            while (arr[i] < pivot)
                  i++;
            while (arr[j] > pivot)
                  j--;
            if (i <= j) {
                  tmp = arr[i];
                  arr[i] = arr[j];
                  arr[j] = tmp;
                  i++;
                  j--;
            }
      };

      return i;
}

void sort(int arr[], int left, int right) {
      int index = partition(arr, left, right);
      if (left < index - 1)
          sort(arr, left, index - 1);
      if (index < right)
          sort(arr, index, right);
}

static int[] a = {6, 10, 13, 5, 8, 3, 2, 11};

public static void main(String[] args) {
    // TODO code application logic here
    QUICKSORT abir=new QUICKSORT();
    System.out.println(Arrays.toString(a));
    abir.sort(a, 0, 7);
    System.out.println(Arrays.toString(a));
}