给定一个单链表 L:L0→L1→…→Ln-1→Ln ,
将其重新排列后变为: L0→Ln→L1→Ln-1→L2→Ln-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.
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。
# 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
/**
* 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;
}
}
}
/**
* 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;
}
}
}
/**
* 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
}
}
/**
* 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;
}
};