【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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水题不解释，一A，代码如下：

/**
 * 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* head;
        if(l1 == NULL)
            return l2;
        if(l2 == NULL)
            return l1;

        if(l1->val < l2->val)
        {
            head = l1;
            l1 = l1->next;
        }
        else
        {
            head = l2;
            l2 = l2->next;
        }
        ListNode* current = head;
        while(l1 != NULL && l2 != NULL){
            if(l1->val < l2->val){
                current->next = l1;
                l1 = l1->next;
                current = current->next;
            }
            else
            {
                current->next = l2;
                l2 = l2->next;
                current = current->next;
            }
        }
        if(l1 != NULL)
            current->next = l1;
        if(l2 != NULL)
            current->next = l2;
        return head;
    }
};

 Java版本代码：

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