【刷题】LC 148. 排序链表

思路：
递归的方式，整个题其实又分成了几个子问题
① 先找到链表中点。将链表拆分
② 递归调用，分别得道左节点、右节点
③ merge左右节点，也就是合并有序链表的思路。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode middle = middleNode(head);
        ListNode rightHead = middle.next;
        middle.next = null;

        ListNode left = sortList(head);
        ListNode right = sortList(rightHead);
        return mergeTwoLists(left, right);
    }

    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
            ListNode sentry = new ListNode(-1);
            ListNode curr = sentry;

            while(l1 != null && l2 != null) {
                if(l1.val < l2.val) {
                    curr.next = l1;
                    l1 = l1.next;
                } else {
                    curr.next = l2;
                    l2 = l2.next;
                }

                curr = curr.next;
            }

            curr.next = l1 != null ? l1 : l2;
            return sentry.next;
    }


    // 中间节点
    public ListNode middleNode(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode slow = head;
        ListNode fast = head.next.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59