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

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1. # Definition for singly-linked list.  
2. # class ListNode(object):  
3. #     def __init__(self, x):  
4. #         self.val = x  
5. #         self.next = None  
6.   
7. class Solution(object):  
8.     def mergeTwoLists(self, l1, l2):  
9.         """ 
10.         :type l1: ListNode 
11.         :type l2: ListNode 
12.         :rtype: ListNode 
13.         """  
14.         n = ListNode(0)  
15.         current = n  
16.         while l1 and l2:  
17.             if l1.val < l2.val:   
18.                 current.next = l1  
19.                 l1 = l1.next  
20.             else:  
21.                 current.next = l2  
22.                 l2 = l2.next  
23.             current = current.next  
24.               
25.         while l1:  
26.             current.next = l1  
27.             l1 = l1.next  
28.             current = current.next  
29.               
30.         while l2:  
31.             current.next = l2  
32.             l2 = l2.next  
33.             current = current.next  
34.               
35.         return n.next  

# 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

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1. /** 
2.  * Definition for singly-linked list. 
3.  * public class ListNode { 
4.  *     int val; 
5.  *     ListNode next; 
6.  *     ListNode(int x) { val = x; } 
7.  * } 
8.  */  
9. public class Solution {  
10.     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {  
11.         ListNode n = new ListNode(0);  
12.         ListNode current = n;  
13.         while(l1 != null && l2 != null){  
14.             if (l1.val < l2.val){  
15.                 current.next = l1;  
16.                 l1 = l1.next;  
17.             }else{  
18.                 current.next = l2;  
19.                 l2 = l2.next;  
20.             }  
21.             current = current.next;  
22.         }  
23.         while (l1 != null){  
24.             current.next = l1;  
25.             l1 = l1.next;   
26.             current = current.next;  
27.         }  
28.         while (l2 != null){  
29.             current.next = l2;  
30.             l2 = l2.next;  
31.             current = current.next;  
32.         }  
33.         return n.next;  
34.     }  
35. }  

/**
 * 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++

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1. /** 
2.  * Definition for singly-linked list. 
3.  * struct ListNode { 
4.  *     int val; 
5.  *     ListNode *next; 
6.  *     ListNode(int x) : val(x), next(NULL) {} 
7.  * }; 
8.  */  
9. class Solution {  
10. public:  
11.     ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {  
12.         ListNode* n = new ListNode(0);  
13.         ListNode* current = n;  
14.         while (l1 != NULL && l2!= NULL){  
15.             if (l1->val < l2->val){  
16.                 current->next = l1;  
17.                 l1 = l1->next;  
18.             }else{  
19.                 current->next = l2;  
20.                 l2 = l2->next;  
21.             }  
22.             current = current->next;  
23.         }  
24.         while(l1 != NULL){  
25.             current->next = l1;  
26.             l1 = l1->next;  
27.             current = current->next;  
28.         }  
29.         while(l2 != NULL){  
30.             current->next = l2;  
31.             l2 = l2->next;  
32.             current = current->next;  
33.         }  
34.         return n->next;  
35.     }  
36. };  

/**
 * 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

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1. /** 
2.  * Definition for singly-linked list. 
3.  * function ListNode(val) { 
4.  *     this.val = val; 
5.  *     this.next = null; 
6.  * } 
7.  */  
8. /** 
9.  * @param {ListNode} l1 
10.  * @param {ListNode} l2 
11.  * @return {ListNode} 
12.  */  
13. var mergeTwoLists = function(l1, l2) {  
14.     var n = new ListNode(0);  
15.     var current = n;  
16.     while(l1 && l2){  
17.         if(l1.val < l2.val) {  
18.             current.next = l1;  
19.             l1 = l1.next;  
20.         }else{  
21.             current.next = l2;  
22.             l2 = l2.next;  
23.         }  
24.         current = current.next;  
25.     }  
26.     while(l1){  
27.         current.next = l1;  
28.         l1 = l1.next;  
29.         current = current.next;  
30.     }  
31.     while(l2){  
32.         current.next = l2;  
33.         l2 = l2.next;  
34.         current = current.next;  
35.     }  
36.     return n.next;  
37. };  

/**
 * 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;
};