Java单链表实现快速排序

普通快排的思路

单链表的实现为

具体函数实现

public void quickSort(ListNode begin, ListNode end) {
//判断为空，判断是不是只有一个节点
if (begin == null || end == null || begin == end)
return;
//从第一个节点和第一个节点的后面一个几点
ListNode first = begin;
ListNode second = begin.next;

int nMidValue = begin.val;
//结束条件，second到最后了
while (second != end.next && second != null) {
if (second.val < nMidValue) {
first = first.next;
//判断一下，避免后面的数比第一个数小，不用换的局面
if (first != second) {
int temp = first.val;
first.val = second.val;
second.val = temp;
}
}
second = second.next;
}
//判断，有些情况是不用换的，提升性能
if (begin != first) {
int temp = begin.val;
begin.val = first.val;
first.val = temp;
}
//前部分递归
quickSort(begin, first);
//后部分递归
quickSort(first.next, end);
}

在main函数是这样的

public static void main(String[] args) {
ListNode head = new ListNode(2);
ListNode l1 = new ListNode(2);
ListNode l2 = new ListNode(5);
ListNode l3 = new ListNode(3);
ListNode l4 = new ListNode(8);
ListNode l5 = new ListNode(4);
ListNode l6 = new ListNode(2);
ListNode l7 = new ListNode(1);

/*ListNode p = l1;
System.out.println(p.equals(head));
System.out.println(p == head);*/

head.next = l1;
l1.next = l2;
l2.next = l3;
l3.next = l4;
l4.next = l5;
l5.next = l6;
l6.next = l7;
l7.next = null;

ListNode p = head;
while (p.next != null) {
System.out.print(p.val);
p = p.next;
}
System.out.print(p.val);
System.out.println();

ListNode begin = head, end = p;
new SingleLinkedListSorting().quickSort(begin, end);

p = head;
while (p != null) {
System.out.print(p.val);
p = p.next;
}

测试结果