Leetcode 148. Sort List 归并排序 in Java

148. Sort List

• Total Accepted: 81218
• Total Submissions: 309907
• Difficulty: Medium

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

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if(head==null || head.next==null) return head;//当分到只有一个node的时候，直接返回

        ListNode mid=findMid(head);

        ListNode secList=mid.next;
        mid.next=null;

        return mergeList(  sortList(head) , sortList(secList)  );
    }

    public ListNode findMid(ListNode head){ //找到中点区分上半段list和下半段list
        ListNode slow=head;
        ListNode fast=head;
        while(fast.next!=null && fast.next.next!=null){ //让slow最终留在前半段list的末尾处，而不是后半段的第一个
            slow=slow.next;
            fast=fast.next.next;
        }
        return slow;
    }

    public ListNode mergeList(ListNode fst,ListNode sec){//合并list

        ListNode newHead=new ListNode(0);
        ListNode tmpHead=newHead;

        while(fst!=null && sec!=null){
            if(fst.val<=sec.val){
                tmpHead.next=fst;
                fst=fst.next;
                tmpHead=tmpHead.next;
                tmpHead.next=null;
            }else{
                tmpHead.next=sec;
                sec=sec.next;
                tmpHead=tmpHead.next;
                tmpHead.next=null;
            }
        }
        tmpHead.next=(fst==null)?sec:fst;
        return newHead.next;
    }
}