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‎Problemset/0143-Reorder List/README.md

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# 143. Reorder List
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You are given the head of a singly linked-list. The list can be represented as:
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L<sub>0</sub> → L<sub>1</sub> → … → L<sub>n - 1</sub> → L<sub>n</sub>
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*Reorder the list to be on the following form:*
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L<sub>0</sub> → L<sub>n</sub> → L<sub>1</sub> → L<sub>n - 1</sub> → L<sub>2</sub> → L<sub>n - 2</sub> → …
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You may not modify the values in the list's nodes. Only nodes themselves may be changed.
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#### Example 1:
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![](https://assets.leetcode.com/uploads/2021/03/04/reorder1linked-list.jpg)
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<pre>
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<strong>Input:</strong> head = [1,2,3,4]
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<strong>Output:</strong> [1,4,2,3]
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</pre>
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#### Example 2:
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![](https://assets.leetcode.com/uploads/2021/03/09/reorder2-linked-list.jpg)
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<pre>
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<strong>Input:</strong> head = [1,2,3,4,5]
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<strong>Output:</strong> [1,5,2,4,3]
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</pre>
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#### Constraints:
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* The number of nodes in the list is in the range <code>[1, 5 * 10<sup>4</sup>]</code>.
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* `1 <= Node.val <= 1000`
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## Solutions (Python)
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### 1. Solution
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```Python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def reorderList(self, head: Optional[ListNode]) -> None:
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"""
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Do not return anything, modify head in-place instead.
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"""
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length = 0
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curr = head
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while curr is not None:
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length += 1
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curr = curr.next
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prev = None
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curr = head
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for _ in range((length + 1) // 2):
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prev = curr
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curr = curr.next
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prev.next = None
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prev = None
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for _ in range(length // 2):
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temp = curr
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curr = curr.next
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temp.next = prev
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prev = temp
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curr0 = head
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curr1 = prev
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for _ in range(length // 2):
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prev0 = curr0
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prev1 = curr1
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curr0 = curr0.next
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curr1 = curr1.next
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prev0.next = prev1
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prev1.next = curr0
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```

‎Problemset/0143-Reorder List/README_CN.md

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# 143. 重排链表
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给定一个单链表 `L` 的头节点 `head` ，单链表 `L` 表示为：
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L<sub>0</sub> → L<sub>1</sub> → … → L<sub>n - 1</sub> → L<sub>n</sub>
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请将其重新排列后变为：
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L<sub>0</sub> → L<sub>n</sub> → L<sub>1</sub> → L<sub>n - 1</sub> → L<sub>2</sub> → L<sub>n - 2</sub> → …
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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。
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#### 示例 1:
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![](https://assets.leetcode.com/uploads/2021/03/04/reorder1linked-list.jpg)
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<pre>
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<strong>输入:</strong> head = [1,2,3,4]
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<strong>输出:</strong> [1,4,2,3]
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</pre>
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#### 示例 2:
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![](https://assets.leetcode.com/uploads/2021/03/09/reorder2-linked-list.jpg)
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<pre>
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<strong>输入:</strong> head = [1,2,3,4,5]
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<strong>输出:</strong> [1,5,2,4,3]
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</pre>
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#### 提示:
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* 链表的长度范围为 <code>[1, 5 * 10<sup>4</sup>]</code>
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* `1 <= Node.val <= 1000`
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## 题解 (Python)
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### 1. 题解
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```Python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def reorderList(self, head: Optional[ListNode]) -> None:
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"""
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Do not return anything, modify head in-place instead.
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"""
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length = 0
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curr = head
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while curr is not None:
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length += 1
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curr = curr.next
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prev = None
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curr = head
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for _ in range((length + 1) // 2):
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prev = curr
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curr = curr.next
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prev.next = None
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prev = None
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for _ in range(length // 2):
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temp = curr
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curr = curr.next
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temp.next = prev
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prev = temp
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curr0 = head
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curr1 = prev
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for _ in range(length // 2):
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prev0 = curr0
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prev1 = curr1
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curr0 = curr0.next
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curr1 = curr1.next
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prev0.next = prev1
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prev1.next = curr0
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```

‎Problemset/0143-Reorder List/Solution.py

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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def reorderList(self, head: Optional[ListNode]) -> None:
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"""
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Do not return anything, modify head in-place instead.
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"""
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length = 0
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curr = head
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while curr is not None:
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length += 1
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curr = curr.next
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prev = None
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curr = head
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for _ in range((length + 1) // 2):
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prev = curr
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curr = curr.next
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prev.next = None
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prev = None
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for _ in range(length // 2):
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temp = curr
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curr = curr.next
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temp.next = prev
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prev = temp
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curr0 = head
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curr1 = prev
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for _ in range(length // 2):
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prev0 = curr0
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prev1 = curr1
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curr0 = curr0.next
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curr1 = curr1.next
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prev0.next = prev1
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prev1.next = curr0

‎README.md

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[139][139l] |[Word Break][139] |![py]&nbsp;&nbsp;![rb]
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[141][141l] |[Linked List Cycle][141] |![py]
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[142][142l] |[Linked List Cycle II][142] |![py]
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[143][143l] |[Reorder List][143] |![py]
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[144][144l] |[Binary Tree Preorder Traversal][144] |![py]
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[145][145l] |[Binary Tree Postorder Traversal][145] |![py]
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[146][146l] |[LRU Cache][146] |![py]
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[139]:Problemset/0139-Word%20Break/README.md#139-word-break
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[141]:Problemset/0141-Linked%20List%20Cycle/README.md#141-linked-list-cycle
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[142]:Problemset/0142-Linked%20List%20Cycle%20II/README.md#142-linked-list-cycle-ii
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[143]:Problemset/0143-Reorder%20List/README.md#143-reorder-list
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[144]:Problemset/0144-Binary%20Tree%20Preorder%20Traversal/README.md#144-binary-tree-preorder-traversal
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[145]:Problemset/0145-Binary%20Tree%20Postorder%20Traversal/README.md#145-binary-tree-postorder-traversal
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[146]:Problemset/0146-LRU%20Cache/README.md#146-lru-cache
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[139l]:https://leetcode.com/problems/word-break/
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[141l]:https://leetcode.com/problems/linked-list-cycle/
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[142l]:https://leetcode.com/problems/linked-list-cycle-ii/
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[143l]:https://leetcode.com/problems/reorder-list/
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[144l]:https://leetcode.com/problems/binary-tree-preorder-traversal/
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[145l]:https://leetcode.com/problems/binary-tree-postorder-traversal/
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[146l]:https://leetcode.com/problems/lru-cache/

‎README_CN.md

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[139][139l] |[单词拆分][139] |![py]&nbsp;&nbsp;![rb]
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[141][141l] |[环形链表][141] |![py]
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[142][142l] |[环形链表 II][142] |![py]
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[143][143l] |[重排链表][143] |![py]
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[144][144l] |[二叉树的前序遍历][144] |![py]
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[145][145l] |[二叉树的后序遍历][145] |![py]
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[146][146l] |[LRU 缓存][146] |![py]
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[139]:Problemset/0139-Word%20Break/README_CN.md#139-单词拆分
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[141]:Problemset/0141-Linked%20List%20Cycle/README_CN.md#141-环形链表
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[142]:Problemset/0142-Linked%20List%20Cycle%20II/README_CN.md#142-环形链表-ii
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[143]:Problemset/0143-Reorder%20List/README_CN.md#143-重排链表
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[144]:Problemset/0144-Binary%20Tree%20Preorder%20Traversal/README_CN.md#144-二叉树的前序遍历
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[145]:Problemset/0145-Binary%20Tree%20Postorder%20Traversal/README_CN.md#145-二叉树的后序遍历
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[146]:Problemset/0146-LRU%20Cache/README_CN.md#146-lru-缓存
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[139l]:https://leetcode.cn/problems/word-break/
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[141l]:https://leetcode.cn/problems/linked-list-cycle/
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[142l]:https://leetcode.cn/problems/linked-list-cycle-ii/
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[143l]:https://leetcode.cn/problems/reorder-list/
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[144l]:https://leetcode.cn/problems/binary-tree-preorder-traversal/
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[145l]:https://leetcode.cn/problems/binary-tree-postorder-traversal/
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[146l]:https://leetcode.cn/problems/lru-cache/

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