143. 重排链表

题目描述

给定一个单链表 L：L₀→L₁→…→L_n-1→L_{n ，}
将其重新排列后变为： L₀→L_n→L₁→L_n-1→L₂→L_n-2→…

你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:

给定链表 1->2->3->4, 重新排列为 1->4->2->3.

示例 2:

给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.

解法

先通过快慢指针找到链表中点，将链表划分为左右两部分。之后反转右半部分的链表，然后将左右两个链接依次连接即可。

Python3

# 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

Java

/**
 * 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;
        }
    }
}

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