Question

谁能看到为什么递归排序示例在第二次调用递归方法后仍继续运行。当第二次通话后i == high_index都在2的位置时，由于某种原因，我在互联网上发现的这段代码再次运行同一行，但具有一些奇怪的值i=4和high_index=6 。我只是看不到错误。所有的递归都应该在第二次之后停止，并且应该对数组进行排序。

查看带有评论的行//WHY DOES THIS LINE RUN TWICE IN A ROW WITH DIFFERENT VALUES?????

package dataClass;

import java.util.Arrays;
public class QuickSort {

    private int temp_array[];
    private int len;

    public void sort(int[] nums) {

        if (nums == null || nums.length == 0) {
            return;
        }
        this.temp_array = nums;
        len = nums.length;
        quickSort(0, len - 1, "First");
        //System.out.println("what");
    }
     private void quickSort(int low_index, int high_index, String one_two) {
        System.out.println("***" + low_index + " " + high_index);
        System.out.println(one_two);
        int i = low_index;
        int j = high_index;
        // calculate pivot number 
        int pivot = temp_array[low_index+(high_index-low_index)/2];
        // Divide into two arrays
        System.out.println(Arrays.toString(temp_array));
        while (i <= j) {
               while (temp_array[i] < pivot) {
                   System.out.println("This happens");
                i++;
            }
            while (temp_array[j] > pivot) {
                System.out.println("Or this happens");
                j--;
            }
            if (i <= j) {
                System.out.println("Execute");

                exchangeNumbers(i, j);
                //move index to next position on both sides
                i++;
                j--;
                System.out.println("i=" + i + " high_index=" + high_index);
            }
        }
        // call quickSort() method recursively

        if (low_index < j) {
            System.out.println("Running 1");
            System.out.println(j + " " + low_index);
            quickSort(low_index, j, "Run 1---------");            
        }


        System.out.println("**i=" + i + " **high_index=" + high_index);  // WHY DOES THIS LINE RUN TWICE IN A ROW WITH DIFFERENT VALUES?????
        System.out.println("Why run again without call to quickSort()?");
        if (i < high_index) {
            System.out.println("Running 2");
            System.out.println(i + " " + high_index);
            quickSort(i, high_index, "Run 2---------");
        }


    }

    private void exchangeNumbers(int i, int j) {
        int temp = temp_array[i];
        temp_array[i] = temp_array[j];
        temp_array[j] = temp;
    }

     // Method to test above
    public static void main(String args[])
    {
        QuickSort ob = new QuickSort();
        int nums[] = {7, -5, 3, 2, 1, 0, 45};
        System.out.println("Original Array:");
        System.out.println(Arrays.toString(nums));
        ob.sort(nums);
        System.out.println("Sorted Array");
        System.out.println(Arrays.toString(nums));
    }
}

Answer 1

尽管我没有真正在这里找到完整的逻辑，但是我相信缺少“ return”语句。当控件出现递归方法调用调用时，该方法将再次执行。但是，一旦执行完递归调用，便会继续执行原始方法，然后您会看到意外的行为！

在进行递归调用时也会在其他地方返回（中断执行流），以防止按预期的方式进一步执行代码块

if (low_index < j) {
            System.out.println("Running 1");
            System.out.println(j + " " + low_index);
            quickSort(low_index, j, "Run 1---------");
            //recursive code invoked, but prevent the control to process downstream code
            return;
        }

递归循环再次运行，而无需在Java中对其进行调用

1 个答案: