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Hello **Programmers/Coders,** Today we are going to share ** solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python**. At Each Problem with Successful submission with

**all Test Cases Passed,**you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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; } pair<ListNode*, ListNode*> merge(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

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