快速排序分区调整

Question

但我想知道如何选择我希望的支点，例如在这个整数列表中，8、7、1、9、11、5、6，我希望选择键6作为我代码中的支点。或者我想选9或者其他什么。我怎样才能把它写进我的代码？非常感谢任何帮助。

package quicksort;

public class quicky {
    private static void quicksort(int[] arr, int left, int right) {
        int index = partition(arr, left, right);
    
        if(left < index - 1)
            quicksort(arr, left, index - 1);
        if(index < right)
            quicksort(arr, index, right);
    }
    private static int partition (int[] arr, int left, int right) {
        int pivot = arr[(left + right) / 2];
        while(left<= right) {
            while(arr[left] < pivot) left++;
            while(arr[right]> pivot) right--;
        
            if(left<= right) {
                int tmp = arr[left];
                arr[left] =arr[right];
                arr[right] = tmp;
            
                left++;
                right--;
            }
        }
        return left;
    }
    public static void main(String[] args) {
        // TODO Auto-generated method stub

        int[] array = new int [] { 8, 7, 1, 9, 11, 5, 6};
        quicksort(array, 0 , array.length-1);
        for(int i = 0; i <array.length; i++)
            System.out.print(array[i]+ " ");        
        }        
    }
}

Answer 1

简单方法：

1. 检查数组中是否存在值。
2. 如果是，则将值与默认数据透视索引中的值交换。
3. 继续使用相同的代码/lomuto分区。

Answer 2

在您的代码中

private static int partition (int[] arr, int left, int right) {
/*
Here in below code only you need to make your changes and that needs to be through out the same as you are calling this from quicksort also you need to make use of the left and right for that, as everytime you will be passing that otherwise it will be static value. 
*/        
int pivot = arr[(left + right) / 2];
        

        while(left<= right) {
            while(arr[left] < pivot) left++;
            while(arr[right]> pivot) right--;
        
            if(left<= right) {
                int tmp = arr[left];
                arr[left] =arr[right];
                arr[right] = tmp;
            
                left++;
                right--;
            }
        }
        return left;
    }