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# Reorder List LeetCode Solution

## Problem – Reorder List LeetCode Solution

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`

## Reorder List LeetCode Solution in Java

``````public void reorderList(ListNode head) {

//Find the middle of the list
while(p2.next!=null&&p2.next.next!=null){
p1=p1.next;
p2=p2.next.next;
}

//Reverse the half after middle  1->2->3->4->5->6 to 1->2->3->6->5->4
ListNode preMiddle=p1;
ListNode preCurrent=p1.next;
while(preCurrent.next!=null){
ListNode current=preCurrent.next;
preCurrent.next=current.next;
current.next=preMiddle.next;
preMiddle.next=current;
}

//Start reorder one by one  1->2->3->6->5->4 to 1->6->2->5->3->4
p2=preMiddle.next;
while(p1!=preMiddle){
preMiddle.next=p2.next;
p2.next=p1.next;
p1.next=p2;
p1=p2.next;
p2=preMiddle.next;
}
}
``````

## Reorder List LeetCode Solution in C++

``````class Solution {
public:
void reorderList(ListNode* head) {
if ((!head) || (!head->next) || (!head->next->next)) return; // Edge cases

stack<ListNode*> my_stack;
ListNode* ptr = head;
int size = 0;
while (ptr != NULL) // Put all nodes in stack
{
my_stack.push(ptr);
size++;
ptr = ptr->next;
}

ListNode* pptr = head;
for (int j=0; j<size/2; j++) // Between every two nodes insert the one in the top of the stack
{
ListNode *element = my_stack.top();
my_stack.pop();
element->next = pptr->next;
pptr->next = element;
pptr = pptr->next->next;
}
pptr->next = NULL;
}
};
``````

## Reorder List LeetCode Solution in Python

``````class Solution:
#step 1: find middle
if not head: return []
while fast.next and fast.next.next:
slow = slow.next
fast = fast.next.next

#step 2: reverse second half
prev, curr = None, slow.next
while curr:
nextt = curr.next
curr.next = prev
prev = curr
curr = nextt
slow.next = None

#step 3: merge lists
``````
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