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) {

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;
}

}

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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