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In this post, you will find the solution for the **Sort List** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Sort List– LeetCode Problem

Sort List– LeetCode Problem

**Problem:**

Given the `head`

of a linked list, return *the list after sorting it in ascending order*.

**Example 1:**

Input:head = [4,2,1,3]Output:[1,2,3,4]

**Example 2:**

Input:head = [-1,5,3,4,0]Output:[-1,0,3,4,5]

**Example 3:**

Input:head = []Output:[]

**Constraints:**

- The number of nodes in the list is in the range
`[0, 5 * 10`

.^{4}] `-10`

^{5}<= Node.val <= 10^{5}

Sort List– LeetCode Solutions

Sort List Solution in C++:

class Solution { public: ListNode* sortList(ListNode* head) { const int length = getLength(head); ListNode dummy(0, head); for (int k = 1; k < length; k *= 2) { ListNode* curr = dummy.next; ListNode* tail = &dummy; while (curr) { ListNode* l = curr; ListNode* r = split(l, k); curr = split(r, k); auto [mergedHead, mergedTail] = merge(l, r); tail->next = mergedHead; tail = mergedTail; } } return dummy.next; } private: int getLength(ListNode* head) { int length = 0; for (ListNode* curr = head; curr; curr = curr->next) ++length; return length; } ListNode* split(ListNode* head, int k) { while (--k && head) head = head->next; ListNode* rest = head ? head->next : nullptr; if (head) head->next = nullptr; return rest; } pairmerge(ListNode* l1, ListNode* l2) { ListNode dummy(0); ListNode* tail = &dummy; while (l1 && l2) { if (l1->val > l2->val) swap(l1, l2); tail->next = l1; l1 = l1->next; tail = tail->next; } tail->next = l1 ? l1 : l2; while (tail->next) tail = tail->next; return {dummy.next, tail}; } };

Sort List Solution in Java:

class Solution { public ListNode sortList(ListNode head) { final int length = getLength(head); ListNode dummy = new ListNode(0, head); for (int k = 1; k < length; k *= 2) { ListNode curr = dummy.next; ListNode tail = dummy; while (curr != null) { ListNode l = curr; ListNode r = split(l, k); curr = split(r, k); ListNode[] merged = merge(l, r); tail.next = merged[0]; tail = merged[1]; } } return dummy.next; } private int getLength(ListNode head) { int length = 0; for (ListNode curr = head; curr != null; curr = curr.next) ++length; return length; } private ListNode split(ListNode head, int k) { while (--k > 0 && head != null) head = head.next; ListNode rest = head == null ? null : head.next; if (head != null) head.next = null; return rest; } private ListNode[] merge(ListNode l1, ListNode l2) { ListNode dummy = new ListNode(0); ListNode tail = dummy; while (l1 != null && l2 != null) { if (l1.val > l2.val) { ListNode temp = l1; l1 = l2; l2 = temp; } tail.next = l1; l1 = l1.next; tail = tail.next; } tail.next = l1 == null ? l2 : l1; while (tail.next != null) tail = tail.next; return new ListNode[] {dummy.next, tail}; } }

Sort List Solution in Python:

class Solution: def sortList(self, head: ListNode) -> ListNode: def split(head: ListNode, k: int) -> ListNode: while k > 1 and head: head = head.next k -= 1 rest = head.next if head else None if head: head.next = None return rest def merge(l1: ListNode, l2: ListNode) -> tuple: dummy = ListNode(0) tail = dummy while l1 and l2: if l1.val > l2.val: l1, l2 = l2, l1 tail.next = l1 l1 = l1.next tail = tail.next tail.next = l1 if l1 else l2 while tail.next: tail = tail.next return dummy.next, tail length = 0 curr = head while curr: length += 1 curr = curr.next dummy = ListNode(0, head) k = 1 while k < length: curr = dummy.next tail = dummy while curr: l = curr r = split(l, k) curr = split(r, k) mergedHead, mergedTail = merge(l, r) tail.next = mergedHead tail = mergedTail k *= 2 return dummy.next