在O(n log n)的时间内使用常数级空间复杂度对链表进行排序。

题目描述
在O(n log n)的时间内使用常数级空间复杂度对链表进行排序。
Sort a linked list in O(n log n) time using constant space complexity.

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if(head == null || head.next == null){
            return head;
        }
        ListNode start = head;
        ListNode mid = getMid(head);
        ListNode end = mid.next;
        mid.next = null;
        return merge(sortList(start),sortList(end));
        
    }
    public ListNode merge(ListNode node1, ListNode node2){
        if(node1 == null) {
            return node2;
        }
        if(node2 == null){
            return node1;
        }
        ListNode min = new ListNode(-1);
        ListNode cur = min;
        while(node1!=null && node2!=null){
            if(node1.val < node2.val){
                cur.next = node1;
                node1 = node1.next;
            }else{
                cur.next = node2;
                node2 = node2.next;
            }
            cur = cur.next;
        }
        if(node1 != null){
            cur.next = node1;
        }
        if(node2 != null ){
            cur.next = node2;
        }
        return min.next;
    }
// 采用快慢指针获取链表的中间节点
    public ListNode getMid(ListNode node){
        if(node == null || node.next == null){
            return node;
        }
        ListNode slow = node;
        ListNode fast = node;
        while(fast.next != null && fast.next.next!= null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
}