题目描述

在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序。

示例 1:

输入: 4->2->1->3
输出: 1->2->3->4

示例 2:

输入: -1->5->3->4->0
输出: -1->0->3->4->5

解法

思路1，归并排序

public ListNode sortList(ListNode head) {if(head == null || head.next==null) {return head;}//获取中间节点ListNode mid = findMid(head);//左右分别递归进行排序ListNode right = sortList(mid.next);mid.next =null;ListNode left = sortList(head);//最终进行mergereturn merge(left, right);
}ListNode merge(ListNode firstNode, ListNode secondNode) {//哑结点ListNode head = new ListNode(0);ListNode cur = head, first = firstNode, second = secondNode;while(first != null && second !=null) {while(first != null && first.val < second.val) {cur.next = first;cur = cur.next;first = first.next;}while(first != null && second !=null && first.val >= second.val) {cur.next = second;cur = cur.next;second = second.next;}}if(first != null) {cur.next = first;}else if (second !=null) {cur.next = second;}return head.next;
}ListNode findMid(ListNode head) {ListNode slowNode = head, fastNode = head.next;while(fastNode != null && fastNode.next!=null) {slowNode = slowNode.next;fastNode = fastNode.next.next;}return slowNode;}