**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:
def split(self, head):
tail1 = head
tail2 = head
while tail1 and tail2 and tail2.next and tail2.next.next:
tail1 = tail1.next
tail2 = tail2.next.next
# now actually break the link!
head2 = tail1.next
tail1.next = None
return head, head2
def reverse(self, head):
if not head or not head.next:
return head
prev, cur = None, head
while cur:
nxt = cur.next
cur.next = prev
prev = cur
cur = nxt
return prev
def interleave(self, head, head2):
resHead = res = LLNode(0)
while head or head2:
if head2:
res.next = head2
res = res.next
head2 = head2.next
if head:
res.next = head
res = res.next
head = head.next
return resHead.next
def solve(self, head):
# break into two halves
head, head2 = self.split(head)
# now reverse the second list ;)
head2 = self.reverse(head2)
# just alternately pick from each
return self.interleave(head, head2)
```

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