# LC 143. Reorder List

### [Problem link](https://leetcode.com/problems/reorder-list/)

###### tags: `leedcode` `python` `c++` `medium` `Linked List`

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


Reorder the list to be on the following form:


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

**Example 1:**
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/04/reorder1linked-list.jpg" style="width: 422px; height: 222px;" />

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

**Example 2:**
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/09/reorder2-linked-list.jpg" style="width: 542px; height: 222px;" />

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 <code>[1, 5 * 10⁴]</code>.
- <code>1 <= Node.val <= 1000</code>

## Solution 1 - Linked List
#### Python

# 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:
        """
        Do not return anything, modify head in-place instead.
        """
        dummy = ListNode(next=head)
        fast = slow = dummy

        # split
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next

        rev = slow.next
        slow.next = None

        # reverse
        node = None
        while rev:
            tmp = rev.next
            rev.next = node
            node = rev
            rev = tmp

        # merge
        while node:
            tmp1, tmp2 = head.next, node.next
            head.next = node
            node.next = tmp1
            head, node = tmp1, tmp2

#### C++ (better than python)

/**
 * 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:
    ListNode* reverse(ListNode *node) {
        ListNode *cur = node;
        ListNode *pre = nullptr;
        while (cur) {
            ListNode *next = cur->next;
            cur->next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }

    void reorderList(ListNode* head) {
        if (head->next == nullptr) {
            return;
        }
        ListNode *fast = head;
        ListNode *pre = nullptr;
        ListNode *slow = head;
        while (fast && fast->next) {
            fast = fast->next->next;
            pre = slow;
            slow = slow->next;
        }
        pre->next = nullptr;
        ListNode *h1 = head;
        ListNode *h2 = reverse(slow);
        while (h1->next) {
            ListNode *next1 = h1->next;
            ListNode *next2 = h2->next;
            h1->next = h2;
            h2->next = next1;
            h1 = next1;
            h2 = next2;
        }
        h1->next = h2;
        return;
    }
};

>### Complexity
>n = The number of nodes in the list
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(1) |

## Note
sol1:
split -> reverse -> merge

ex1:
1 -> 2 -> 3 -> 4 -> 5 -> 6 -> None

split:
1 -> 2 -> 3 -> None
4 -> 5 -> 6 -> None

反轉:
4 -> 5 -> 6 -> None => 6 -> 5 -> 4 -> None

合併:
1 -> 6 -> 2 -> 5 -> 3 -> 4 -> None

ex2:
1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> None

split:
1 -> 2 -> 3 -> 4 -> None
5 -> 6 -> 7 -> None

反轉:
5 -> 6 -> 7 -> None => 7 -> 6 -> 5 -> None

合併:
1 -> 7 -> 2 -> 6 -> 3 -> 5 -> 4 -> None