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#
Solution 1#
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| # 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: Optional[ListNode]) -> None:
# 快慢指针找到链表中点
fast = slow = head
while fast.next and fast.next.next:
slow = slow.next
fast = 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
# 合并
while pre:
t = pre.next
pre.next = cur.next
cur.next = pre
cur, pre = pre.next, t
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| /**
* 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) {
// 快慢指针找到链表中点
ListNode fast = head, slow = head;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
// cur 指向右半部分链表
ListNode cur = slow.next;
slow.next = null;
// 反转右半部分链表
ListNode pre = null;
while (cur != null) {
ListNode 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
// 合并
while (pre != null) {
ListNode t = pre.next;
pre.next = cur.next;
cur.next = pre;
cur = pre.next;
pre = t;
}
}
}
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| /**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
void reorderList(ListNode* head) {
// 快慢指针找到链表中点
ListNode* fast = head;
ListNode* slow = head;
while (fast->next && fast->next->next) {
slow = slow->next;
fast = fast->next->next;
}
// cur 指向右半部分链表
ListNode* cur = slow->next;
slow->next = nullptr;
// 反转右半部分链表
ListNode* pre = nullptr;
while (cur) {
ListNode* 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
// 合并
while (pre) {
ListNode* t = pre->next;
pre->next = cur->next;
cur->next = pre;
cur = pre->next;
pre = t;
}
}
};
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| /**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reorderList(head *ListNode) {
// 快慢指针找到链表中点
fast, slow := head, head
for fast.Next != nil && fast.Next.Next != nil {
slow, fast = slow.Next, fast.Next.Next
}
// cur 指向右半部分链表
cur := slow.Next
slow.Next = nil
// 反转右半部分链表
var pre *ListNode
for cur != nil {
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
// 合并
for pre != nil {
t := pre.Next
pre.Next = cur.Next
cur.Next = pre
cur, pre = pre.Next, t
}
}
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| /**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
/**
Do not return anything, modify head in-place instead.
*/
function reorderList(head: ListNode | null): void {
const arr = [];
let node = head;
while (node.next != null) {
arr.push(node);
node = node.next;
}
let l = 0;
let r = arr.length - 1;
while (l < r) {
const start = arr[l];
const end = arr[r];
[end.next.next, start.next, end.next] = [start.next, end.next, null];
l++;
r--;
}
}
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| // Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
// pub val: i32,
// pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
// #[inline]
// fn new(val: i32) -> Self {
// ListNode {
// next: None,
// val
// }
// }
// }
use std::collections::VecDeque;
impl Solution {
pub fn reorder_list(head: &mut Option<Box<ListNode>>) {
let mut tail = &mut head.as_mut().unwrap().next;
let mut head = tail.take();
let mut deque = VecDeque::new();
while head.is_some() {
let next = head.as_mut().unwrap().next.take();
deque.push_back(head);
head = next;
}
let mut flag = false;
while !deque.is_empty() {
*tail = if flag { deque.pop_front().unwrap() } else { deque.pop_back().unwrap() };
tail = &mut tail.as_mut().unwrap().next;
flag = !flag;
}
}
}
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| /**
* 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) {
// 快慢指针找到链表中点
let slow = head;
let fast = head;
while (fast.next && fast.next.next) {
slow = slow.next;
fast = fast.next.next;
}
// cur 指向右半部分链表
let cur = slow.next;
slow.next = null;
// 反转右半部分链表
let pre = null;
while (cur) {
const 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
// 合并
while (pre) {
const t = pre.next;
pre.next = cur.next;
cur.next = pre;
cur = pre.next;
pre = t;
}
};
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| /**
* 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) {
// 快慢指针找到链表中点
ListNode slow = head;
ListNode fast = head;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
// cur 指向右半部分链表
ListNode cur = slow.next;
slow.next = null;
// 反转右半部分链表
ListNode pre = null;
while (cur != null) {
ListNode 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
// 合并
while (pre != null) {
ListNode t = pre.next;
pre.next = cur.next;
cur.next = pre;
cur = pre.next;
pre = t;
}
}
}
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Solution 2#
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| /**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
/**
Do not return anything, modify head in-place instead.
*/
function reorderList(head: ListNode | null): void {
let slow = head;
let fast = head;
// 找到中心节点
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
// 反转节点
let next = slow.next;
slow.next = null;
while (next != null) {
[next.next, slow, next] = [slow, next, next.next];
}
// 合并
let left = head;
let right = slow;
while (right.next != null) {
const next = left.next;
left.next = right;
right = right.next;
left.next.next = next;
left = left.next.next;
}
}
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