LettCode—sort-list（将链表排序）—Java

题目描述：

Sort a linked list in O(n log n) time using constant space complexity.

对链表进行归并排序

思路解析：

• 归并排序的思想：比较大小，把小的先进行排序，剩下的再直接放入。
• 找到链表中间点的方法，也就是分为两个list：firstlist和secondlist
• 递归分解两个list，然后Merge
• Merge的过程需要新建节点：ListNode fakelist = new ListNode(-1);ListNode curr = fakelist;

代码：

/**
 * 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 slow=head,fast=head,firstlist=head;
        while(fast.next!=null&&fast.next.next!=null){
            slow=slow.next;
            fast=fast.next.next;
        }
        ListNode secondlist = slow.next;
        slow.next=null;
        return mergeTwoLists(sortList(firstlist),sortList(secondlist));
    }
    public ListNode mergeTwoLists(ListNode leftlist,ListNode rightlist){
        if(leftlist == null)
            return null;
        if(rightlist == null)
            return null;
        ListNode fakehead = new ListNode(-1);
        ListNode curr = fakehead;
        while(leftlist!=null && rightlist!=null){
            if(leftlist.val<=rightlist.val){
                curr.next = leftlist;
                leftlist=leftlist.next;
            }else{
                curr.next = rightlist;
                rightlist = rightlist.next;
            }
            curr= curr.next;
        }
        curr.next = leftlist!=null?leftlist:rightlist;
        return fakehead.next;
    }
}