143. 重排链表

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reorderList(self, head: ListNode) -> None:
        """
        Do not return anything, modify head in-place instead.
        """
        if head is None or head.next is None:
            return
        slow, fast = head, head.next
        # 快慢指针找到链表中点
        while fast and fast.next:
            slow, fast = slow.next, fast.next.next
        cur = slow.next
        slow.next = None
        pre = None
        # cur 指向右半部分的链表，反转
        while cur:
            t = cur.next
            cur.next = pre
            pre = cur
            cur = t
        cur = head

        # 将左右链表依次连接
        while pre:
            t1 = cur.next
            cur.next = pre
            cur = t1
            t2 = pre.next
            pre.next = t1
            pre = t2

/**
 * 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 || head.next == null) {
            return;
        }
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode cur = slow.next;
        slow.next = null;
        ListNode pre = null;
        while (cur != null) {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        cur = head;
        while (pre != null) {
            ListNode t1 = cur.next;
            cur.next = pre;
            cur = t1;
            ListNode t2 = pre.next;
            pre.next = cur;
            pre = t2;
        }
    }
}