Merge Two Sorted Lists

Problem Description
Given 2 Sorted LinkedList, merge them and return a new Sorted LinkedList.
Solution
This problem is a typical operation used in Merge Sort algorithm.
I. Python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        n = ListNode(0)
        current = n
        while l1 and l2:
            if l1.val < l2.val: 
                current.next = l1
                l1 = l1.next
            else:
                current.next = l2
                l2 = l2.next
            current = current.next
            
        while l1:
            current.next = l1
            l1 = l1.next
            current = current.next
            
        while l2:
            current.next = l2
            l2 = l2.next
            current = current.next
            
        return n.next

II. Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode n = new ListNode(0);
        ListNode current = n;
        while(l1 != null && l2 != null){
            if (l1.val < l2.val){
                current.next = l1;
                l1 = l1.next;
            }else{
                current.next = l2;
                l2 = l2.next;
            }
            current = current.next;
        }
        while (l1 != null){
            current.next = l1;
            l1 = l1.next; 
            current = current.next;
        }
        while (l2 != null){
            current.next = l2;
            l2 = l2.next;
            current = current.next;
        }
        return n.next;
    }
}

III. C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* n = new ListNode(0);
        ListNode* current = n;
        while (l1 != NULL && l2!= NULL){
            if (l1->val < l2->val){
                current->next = l1;
                l1 = l1->next;
            }else{
                current->next = l2;
                l2 = l2->next;
            }
            current = current->next;
        }
        while(l1 != NULL){
            current->next = l1;
            l1 = l1->next;
            current = current->next;
        }
        while(l2 != NULL){
            current->next = l2;
            l2 = l2->next;
            current = current->next;
        }
        return n->next;
    }
};

IV. Javascript

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var mergeTwoLists = function(l1, l2) {
    var n = new ListNode(0);
    var current = n;
    while(l1 && l2){
        if(l1.val < l2.val) {
            current.next = l1;
            l1 = l1.next;
        }else{
            current.next = l2;
            l2 = l2.next;
        }
        current = current.next;
    }
    while(l1){
        current.next = l1;
        l1 = l1.next;
        current = current.next;
    }
    while(l2){
        current.next = l2;
        l2 = l2.next;
        current = current.next;
    }
    return n.next;
};