**Sort a Linked List - Amazon Top Interview Questions**

### Problem Statement :

Given a singly linked list of integers node, sort the nodes by their values in ascending order. Constraints n ≤ 100,000 where n is the number of nodes in node Example 1 Input node = [14, 13, 10] Output [10, 13, 14]

### Solution :

` ````
Solution in C++ :
LLNode* merge(LLNode* left, LLNode* right) {
LLNode dummy_head;
LLNode* curr = &dummy_head;
while (left && right) {
if (left->val < right->val) {
curr->next = left;
left = left->next;
curr = curr->next;
curr->next = nullptr;
} else {
curr->next = right;
right = right->next;
curr = curr->next;
curr->next = nullptr;
}
}
while (left) {
curr->next = left;
left = left->next;
curr = curr->next;
curr->next = nullptr;
}
while (right) {
curr->next = right;
right = right->next;
curr = curr->next;
curr->next = nullptr;
}
return dummy_head.next;
}
LLNode* merge_sort(LLNode* node) {
if (!node || !node->next) return node;
LLNode *slow = node, *fast = node, *prev = nullptr;
while (fast && fast->next) {
prev = slow;
slow = slow->next;
fast = fast->next->next;
}
prev->next = nullptr;
LLNode* left = merge_sort(node);
LLNode* right = merge_sort(slow);
return merge(left, right);
}
LLNode* solve(
LLNode* node) { // Merge sort takes O(N logN) time and O(log N) for recursive call stack
return merge_sort(node);
}
```

` ````
Solution in Python :
# class LLNode:
# def __init__(self, val, next=None):
# self.val = val
# self.next = next
class Solution:
def solve(self, node):
if node is None or node.next is None:
return node
slow, fast = node, node
while fast is not None and fast.next is not None:
prev = slow
slow = slow.next
fast = fast.next.next
prev.next = None
l1 = self.solve(node)
l2 = self.solve(slow)
return self.mergeTwoLists(l1, l2)
def mergeTwoLists(self, l1, l2):
if l1 is None and l2 is not None:
return l2
elif l1 is not None and l2 is None:
return l1
elif l1 is None and l2 is None:
return None
t1 = l1
prev = None # Can add a dummy node to avoid prev=None edge case handling
while t1 is not None and l2 is not None:
if l2.val <= t1.val:
if prev is None:
save = l2
l2 = l2.next
save.next = t1
l1 = save
prev = save
else:
save = l2
l2 = l2.next
prev.next = save
save.next = t1
prev = save
else:
prev = t1
t1 = t1.next
if t1 is None and l2 is not None:
prev.next = l2
return l1
```

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