【坚持每日一题10.25】重排链表

给定一个单链表 L 的头节点 head ，单链表 L 表示为：

L0 → L1 → … → Ln-1 → Ln 
请将其重新排列后变为：

L0 → Ln → L1 → Ln-1 → L2 → Ln-2 → …

不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例 1:

输入: head = [1,2,3,4]
输出: [1,4,2,3]
示例 2:

输入: head = [1,2,3,4,5]
输出: [1,5,2,4,3]

java代码：

/**
 * 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 void reorderList(ListNode head) {
        if (head == null) {
            return;
        }
        ListNode mid = middleNode(head);
        ListNode l1 = head;
        ListNode l2 = mid.next;
        mid.next = null;
        l2 = reverseList(l2);
        mergeList(l1, l2);
    }

    public ListNode middleNode(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }

    public ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode nextTemp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextTemp;
        }
        return prev;
    }

    public void mergeList(ListNode l1, ListNode l2) {
        ListNode l1_tmp;
        ListNode l2_tmp;
        while (l1 != null && l2 != null) {
            l1_tmp = l1.next;
            l2_tmp = l2.next;

            l1.next = l2;
            l1 = l1_tmp;

            l2.next = l1;
            l2 = l2_tmp;
        }
    }
}
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
60
61