# Back to Front Linked List - Amazon Top Interview Questions

### Problem Statement :

Given a singly linked list node, reorder it such that we take: the last node, and then the first node, and then the second last node, and then the second node, etc.

Can you do it in \mathcal{O}(1)O(1) space?

Constraints

0 ≤ n ≤ 100,000 where n is the number of nodes in node

Example 1

Input

node = [0, 1, 2, 3]

Output

[3, 0, 2, 1]

### Solution :

                        Solution in C++ :

LLNode* solve(LLNode* node) {
LLNode* slow = node;
LLNode* fast = node;
while (fast) {
fast = fast->next;
if (fast) {
fast = fast->next;
slow = slow->next;
}
}
LLNode* pre = NULL;
LLNode* cur = slow;
LLNode* nex;
while (cur) {
nex = cur->next;
cur->next = pre;
pre = cur;
cur = nex;
}
LLNode* temp = pre;
while (temp && node) {
LLNode* nex1 = temp->next;
LLNode* nex2 = node->next;
temp->next = node;
node->next = nex1;
temp = nex1;
node = nex2;
}
return pre;
}


                        Solution in Java :

import java.util.*;

/**
* class LLNode {
*   int val;
*   LLNode next;
* }
*/
class Solution {
public LLNode solve(LLNode node) {
LLNode fast = node, slow = node;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
LLNode left = node, right = reverse(slow);
LLNode prev = new LLNode();
LLNode dummy = prev;
while (left != null && right != null) {
prev.next = right;
right = right.next;
prev.next.next = left;
left = left.next;
prev = prev.next.next;
}
prev.next = null;
return dummy.next;
}

private LLNode reverse(LLNode node) {
LLNode prev = null, curr = node, next;
while (curr != null) {
next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
return prev;
}
}


                        Solution in Python :

class Solution:

while tail1 and tail2 and tail2.next and tail2.next.next:
tail1 = tail1.next
tail2 = tail2.next.next

# now actually break the link!
tail1.next = None

while cur:
nxt = cur.next
cur.next = prev
prev = cur
cur = nxt

return prev

res = res.next

res = res.next

# break into two halves

# now reverse the second list ;)

# just alternately pick from each
return self.interleave(head, head2)


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