README_EN.md

143. Reorder List

中文文档

Description

You are given the head of a singly linked-list. The list can be represented as:

L0 → L1 → … → Ln - 1 → Ln

Reorder the list to be on the following form:

L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …

You may not modify the values in the list's nodes. Only nodes themselves may be changed.

 

Example 1:

Input: head = [1,2,3,4]
Output: [1,4,2,3]

Example 2:

Input: head = [1,2,3,4,5]
Output: [1,5,2,4,3]

 

Constraints:

• The number of nodes in the list is in the range [1, 5 * 104].
• 1 <= Node.val <= 1000

Solutions

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
        while fast and fast.next:
            slow, fast = slow.next, fast.next.next

        cur = slow.next
        slow.next = None

        pre = None
        while cur:
            t = cur.next
            cur.next = pre
            pre, cur = cur, t
        cur = head

        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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