21_Merge_Two_Sorted_Lists

Leetcode 21: Merge Two Sorted Lists
Detailed video explanation: https://youtu.be/yn6kTAkf9Mc
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C++:
----
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if(!l1) return l2;
        if(!l2) return l1;
        
        ListNode* head = l1;
        if(l1->val > l2->val){
            head = l2;
            l2 = l2->next;
        } else
            l1 = l1->next;
        
        ListNode* curr = head;
        while(l1 && l2){
            if(l1->val < l2->val){
                curr->next = l1;
                l1 = l1->next;
            } else {
                curr->next = l2;
                l2 = l2->next;
            }
            curr = curr->next;
        }
        if(!l1) curr->next = l2;
        else curr->next = l1;
        
        return head;
    }
};



Java:
-----
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null) return l2;
        if(l2 == null) return l1;
        
        ListNode head = l1;
        if(l1.val > l2.val){
            head = l2;
            l2 = l2.next;
        } else
            l1 = l1.next;
        
        ListNode curr = head;
        while(l1 != null && l2 != null){
            if(l1.val < l2.val){
                curr.next = l1;
                l1 = l1.next;
            } else {
                curr.next = l2;
                l2 = l2.next;
            }
            curr = curr.next;
        }
        if(l1 == null) curr.next = l2;
        else curr.next = l1;
        
        return head;
    }
}




Python3:
--------
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        if l1 == None: return l2
        if l2 == None: return l1
        
        head = l1
        if l1.val > l2.val:
            head = l2
            l2 = l2.next
        else:
            l1 = l1.next
        
        curr = head
        while l1 != None and l2 != None:
            if l1.val < l2.val:
                curr.next = l1
                l1 = l1.next
            else:
                curr.next = l2
                l2 = l2.next
            curr = curr.next
        
        if l1 == None: curr.next = l2
        else: curr.next = l1
        
        return head