README.md

143. 重排链表

English Version

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

        # 快慢指针找到链表中点
        slow, fast = head, head.next
        while fast and fast.next:
            slow, fast = slow.next, fast.next.next

        # cur 指向右半部分链表
        cur = slow.next
        slow.next = None

        # 反转右半部分链表
        pre = None
        while cur:
            t = cur.next
            cur.next = pre
            pre, cur = cur, t
        cur = head
        # 此时 cur, pre 分别指向链表左右两半的第一个节点

        while pre:
            t = pre.next
            pre.next = cur.next
            cur.next = pre
            cur, pre = pre.next, t

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 t = pre.next;
            pre.next = cur.next;
            cur.next = pre;
            cur = pre.next;
            pre = t;
        }
    }
}

C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public 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 t = pre.next;
            pre.next = cur.next;
            cur.next = pre;
            cur = pre.next;
            pre = t;
        }
    }
}

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
 func reorderList(head *ListNode)  {
    if head == nil || head.Next == nil {
        return
    }
    slow, fast := head, head.Next
    for fast != nil && fast.Next != nil {
        slow, fast = slow.Next, fast.Next.Next
    }

    cur := slow.Next
    slow.Next = nil

    var pre *ListNode
    for cur != nil {
        t := cur.Next
        cur.Next = pre
        pre, cur = cur, t
    }
    cur = head

    for pre != nil {
        t := pre.Next
        pre.Next = cur.Next
        cur.Next = pre
        cur, pre = pre.Next, t
    }
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {void} Do not return anything, modify head in-place instead.
 */
var reorderList = function (head) {
    if (!head || !head.next) {
        return;
    }
    let slow = head;
    let fast = head.next;
    while (fast && fast.next) {
        slow = slow.next;
        fast = fast.next.next;
    }

    let cur = slow.next;
    slow.next = null;

    let pre = null;
    while (cur) {
        const t = cur.next;
        cur.next = pre;
        pre = cur;
        cur = t;
    }
    cur = head;

    while (pre) {
        const t = pre.next;
        pre.next = cur.next;
        cur.next = pre;
        cur = pre.next;
        pre = t;
    }
};

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