我在github中找到了quickselect算法的代码，也就是order-statistics算法。这个代码工作正常。
我不理解medianOf3方法，它应该按排序顺序排列第一个、中间和最后一个索引但实际上，在调用medianof3方法之后，当我对数组进行分组时，它不会这样做。
除了最后一次调用swap(list, centerIndex, rightIndex - 1);之外，我可以按照这个方法来确定它在做什么。有人能解释一下为什么叫这个吗？
import java.util.Arrays;
/**
* This program determines the kth order statistic (the kth smallest number in a
* list) in O(n) time in the average case and O(n^2) time in the worst case. It
* achieves this through the Quickselect algorithm.
*
* @author John Kurlak <[email protected]>
* @date 1/17/2013
*/
public class Quickselect {
/**
* Runs the program with an example list.
*
* @param args The command-line arguments.
*/
public static void main(String[] args) {
int[] list = { 3, 5, 9, 10, 7, 40, 23, 45, 21, 2 };
int k = 6;
int median = medianOf3(list, 0, list.length-1);
System.out.println(median);
System.out.println("list is "+ Arrays.toString(list));
Integer kthSmallest = quickselect(list, k);
if (kthSmallest != null) {
System.out.println("The kth smallest element in the list where k=" + k + " is " + kthSmallest + ".");
} else {
System.out.println("There is no kth smallest element in the list where k=" + k + ".");
}
System.out.println(Arrays.toString(list));
}
/**
* Determines the kth order statistic for the given list.
*
* @param list The list.
* @param k The k value to use.
* @return The kth order statistic for the list.
*/
public static Integer quickselect(int[] list, int k) {
return quickselect(list, 0, list.length - 1, k);
}
/**
* Recursively determines the kth order statistic for the given list.
*
* @param list The list.
* @param leftIndex The left index of the current sublist.
* @param rightIndex The right index of the current sublist.
* @param k The k value to use.
* @return The kth order statistic for the list.
*/
public static Integer quickselect(int[] list, int leftIndex, int rightIndex, int k) {
// Edge case
if (k < 1 || k > list.length) {
return null;
}
// Base case
if (leftIndex == rightIndex) {
return list[leftIndex];
}
// Partition the sublist into two halves
int pivotIndex = randomPartition(list, leftIndex, rightIndex);
int sizeLeft = pivotIndex - leftIndex + 1;
// Perform comparisons and recurse in binary search / quicksort fashion
if (sizeLeft == k) {
return list[pivotIndex];
} else if (sizeLeft > k) {
return quickselect(list, leftIndex, pivotIndex - 1, k);
} else {
return quickselect(list, pivotIndex + 1, rightIndex, k - sizeLeft);
}
}
/**
* Randomly partitions a set about a pivot such that the values to the left
* of the pivot are less than or equal to the pivot and the values to the
* right of the pivot are greater than the pivot.
*
* @param list The list.
* @param leftIndex The left index of the current sublist.
* @param rightIndex The right index of the current sublist.
* @return The index of the pivot.
*/
public static int randomPartition(int[] list, int leftIndex, int rightIndex) {
int pivotIndex = medianOf3(list, leftIndex, rightIndex);
int pivotValue = list[pivotIndex];
int storeIndex = leftIndex;
swap(list, pivotIndex, rightIndex);
for (int i = leftIndex; i < rightIndex; i++) {
if (list[i] <= pivotValue) {
swap(list, storeIndex, i);
storeIndex++;
}
}
swap(list, rightIndex, storeIndex);
return storeIndex;
}
/**
* Computes the median of the first value, middle value, and last value
* of a list. Also rearranges the first, middle, and last values of the
* list to be in sorted order.
*
* @param list The list.
* @param leftIndex The left index of the current sublist.
* @param rightIndex The right index of the current sublist.
* @return The index of the median value.
*/
public static int medianOf3(int[] list, int leftIndex, int rightIndex) {
int centerIndex = (leftIndex + rightIndex) / 2;
if (list[leftIndex] > list[rightIndex]) {
swap(list, leftIndex, centerIndex);
}
if (list[leftIndex] > list[rightIndex]) {
swap(list, leftIndex, rightIndex);
}
if (list[centerIndex] > list[rightIndex]) {
swap(list, centerIndex, rightIndex);
}
swap(list, centerIndex, rightIndex - 1);
return rightIndex - 1;
}
/**
* Swaps two elements in a list.
*
* @param list The list.
* @param index1 The index of the first element to swap.
* @param index2 The index of the second element to swap.
*/
public static void swap(int[] list, int index1, int index2) {
int temp = list[index1];
list[index1] = list[index2];
list[index2] = temp;
}
}
最佳答案
所以我写了原始代码，但是我做了很差的工作使它可读。
回头看，我不认为这行代码是必要的，但我认为这是一个小的优化。如果我们删除代码行并返回centerIndex，它似乎可以正常工作，没有任何问题。
不幸的是，它执行的优化应该从medianOf3()中重构出来并移到randomPartition()中。
从本质上讲，优化是我们希望在分区子阵列之前尽可能“部分排序”。原因是：我们的数据排序越多，我们未来的分区选择就越好，这意味着我们的运行时间有望接近O（n）而不是O（n^2）在randomPartition()方法中，我们将轴值移动到我们正在查看的子阵列的最右侧。这会将最右边的值移到子数组的中间这是不需要的，因为最右边的值应该是“较大的值”。我的代码试图通过将轴索引放在最右边的索引旁边来防止这种情况然后，当轴索引与randomPartition()中最右边的索引交换时，最右边的“较大”值不会移动到子数组的中间，而是保持在右边附近。