LeetCode: Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

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The solution is trivial. Just keep in mind to deal with every corner case. Find the new head first and then give each list a iterator to iterate all elements.

C++ Code:

/*
 * func: merge_sorted_list
 * goal: merget two sorted list to one list
 * @param l1: list 1
 * @param l2: list 2
 * return: new head of the merged list
 */
ListNode* merge_sorted_list(ListNode *l1, ListNode *l2){
    if(!l1){
        return l2;
    }
    if(!l2){
        return l1;
    }
    
    ListNode *it1 = l1;
    ListNode *it2 = l2;
    ListNode *new_head = nullptr;
    //Set up new head
    if(it1->val >= it2->val){
        new_head = it2;
        it2 = it2->next;
    }else{
        new_head = it1;
        it1 = it1->next;
    }
    
    ListNode *it = new_head;
    while(it1 && it2){
        if(it1->val <= it2->val){
            it->next = it1;
            it1 = it1->next;
            it = it->next;
        }else{
            it->next = it2;
            it2 = it2->next;
            it = it->next;
        }
    }
    if(it1){
        it->next = it1;
    }
    if(it2){
        it->next = it2;
    }
    
    return new_head;
}

Python Code:

# func: merge two sorted list into one
# @param l1: list 1
# @param l2: list 2
# @return: the head of the new list
def merge_sorted_list(l1, l2):
    if not l1:
        return l2
    if not l2:
        return l1

    iter1 = l1
    iter2 = l2
    # Set up the new head node
    if l1.val <= l2.val:
        new_head = l1
        iter1 = iter1.next
    else:
        new_head = l2
        iter2 = iter2.next
    iter = new_head
    while iter1 and iter2:
        if iter1.val <= iter2.val:
            iter.next = iter1
            iter1 = iter1.next
            iter = iter.next
        else:
            iter.next = iter2
            iter2 = iter2.next
            iter = iter.next
    #Contatenate remaining part
    if iter1:
        iter.next = iter1
    if iter2:
        iter.next = iter2

    return new_head