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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 * 104]. -105 <= Node.val <= 105
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 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