[英]Doubly Linked List QuickSort Implementation Problem
我已经实现了一个经典的双向链表:
class Node<T> {
protected T data;
protected Node<T> next, prev;
}
class DoublyLinkedList<T extends Comparable<T>> {
protected Node<T> front;
protected Node<T> back;
protected int size;
// methods
}
现在为了能够对其进行排序,我添加了以下实现经典快速排序算法的方法:
public void sort(Comparator<T> comparator) {
quickSort(front, back, comparator);
}
private void quickSort(Node<T> begin, Node<T> end, Comparator<T> comparator) {
if (end != null && begin != end && begin != end.next) {
var temp = partition(begin, end, comparator);
quickSort(begin, temp.prev, comparator);
quickSort(temp.next, end, comparator);
}
}
private Node<T> partition(Node<T> begin, Node<T> end, Comparator<T> comparator) {
var pivot = end.data;
var i = begin.prev;
Node<T> next;
for (var j = begin; j != end; j = next) {
next = j.next;
if (comparator.compare(j.data, pivot) < 0) {
i = (i == null) ? begin : i.next;
swapData(i, j);
}
}
i = (i == null) ? begin : i.next;
swapData(i, end);
return i;
}
private void swapData(Node<T> a, Node<T> b) {
var temp = a.data;
a.data = b.data;
b.data = temp;
}
上面的代码产生了正确的结果,但是,我决定交换节点而不是数据,所以我介绍了这些方法:
private void swapNodes(Node<T> a, Node<T> b) {
if (a == b) return;
if (a == null || b == null) {
throw new NullPointerException();
}
if (a.next == b) {
var before = a.prev;
var after = b.next;
link(before, b);
link(b, a);
link(a, after);
} else if (b.next == a) {
var before = b.prev;
var after = a.next;
link(before, a);
link(a, b);
link(b, after);
} else {
var aPrev = a.prev;
var aNext = a.next;
var bPrev = b.prev;
var bNext = b.next;
link(aPrev, b);
link(b, aNext);
link(bPrev, a);
link(a, bNext);
}
}
private void link(Node<T> a, Node<T> b) {
if (a != null)
a.next = b;
else
front = b;
if (b != null)
b.prev = a;
else
back = a;
}
并将这些更改添加到partition
方法中:
private Node<T> partition(Node<T> begin, Node<T> end, Comparator<T> comparator) {
var pivot = end.data;
var i = begin.prev;
Node<T> next;
for (var j = begin; j != end; j = next) {
next = j.next;
if (comparator.compare(j.data, pivot) < 0) {
i = (i == null) ? begin : i.next;
//swapData(i, j);
swapNodes(i, j);
i = j;
}
}
i = (i == null) ? begin : i.next;
//swapData(i, end);
swapNodes(i, end);
//return i;
return end;
}
此时代码无法正常工作,我不知道为什么。 我错过了什么?
编辑:
预期的 output 是排序后的输入,而在第二种情况下则不是。
例子:
Initial :[2, 9, 8, 3, 6, 2, 4, 1, 7, 6]
Expected:[1, 2, 2, 3, 4, 6, 6, 7, 8, 9]
Actual: [1, 3, 2, 4, 2, 6, 9, 6, 7, 8]
可以在此处找到一个工作示例: https://ideone.com/UQrzY1
编辑2:
提供了一个较短的示例和输入/输出。
“交换节点变体”中的错误很难确定是有原因的:
你不支持调试。 养成让类提供基本toString()
的习惯:
/** doubly linked list node */
static class Node<T> {
…
/** constructs a <code>Node</code> given data, next & prev */
public Node(T d, Node…
@Override
public String toString() {
return String.valueOf(data);
}
}
列表有点复杂 -
/** Append string representations of <code>node</code>s
* <code>data</code> to <code>head</code>, following
* <code>next</code>s til <code>end</code> (or <code>null</code>)
* (inclusive)
*/
Appendable append(Node<T> node, final Node<T> end,
CharSequence separator, Appendable head) {
try {
while (end != node) {
head.append(String.valueOf(node));
if (null == node
|| null == (node = node.next) && null == end)
return head;
head.append(separator);
}
head.append(String.valueOf(node));
} catch (IOException e) {
e.printStackTrace();
}
return head;
}
@Override
public String toString() {
return ((StringBuilder)append(front, null, ", ",
new StringBuilder("["))).append(']').toString();
}
void bug(String label, Node<T> node, final Node<T> end) {
System.out.append(((StringBuilder)append(node, end, ", ",
new StringBuilder(label).append('('))).append(")\n"));
}
String verbose(Node<T> n) {
return "+" + n.prev + "<-" + n + "->" + n.next;
}
private void quickSort(Node<T> begin, Node<T> end, Comparator<T> comparator) {
bug("quicksort", begin, end);
if (end != null && begin != end && begin != end.next) {
Node<T> temp = partition(begin, end, comparator);
System.out.println("begin: " + begin + ", temp: "
+ verbose(temp) + ", temp == end: " + (temp == end));
quickSort(begin, temp.prev, comparator);
bug("between", begin, temp.prev);
quickSort(temp.next, end, comparator);
}
}
使用上面的侵入式调试,您可以看到end
不会停留在正确部分的末尾 - 如何在 Lomuto 分区中选择 pivot 元素。
begin
也不会停留在左侧部分的开头 - 您似乎分别需要begin
的前任和end
的后继者的前任。
在列表之前和之后没有哨兵节点的情况下会出现一大堆特殊情况。
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