leetcode 21. Merge Two Sorted Lists

题目描述：
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.

很简单就是一个归并两个已经排好序的两个链表。也是归并排序中核心的代码。主要用到了链表的知识，考验了链表的构建。
下面是解答：

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if (l1 == NULL&&l2 == NULL) {
            return NULL;
        }
        ListNode* p = new ListNode(0);
        ListNode* result = p;
        ListNode* rl1 = l1;
        ListNode* rl2 = l2;
        while (rl1 != NULL || rl2 != NULL) {
            //当有一个链表已经为空
            if (rl1 == NULL) {
                p->next = rl2;
                break;
            }
            else if (rl2 == NULL) {
                p->next= rl1;
                break;
            }
            //每次获取较小的那个值来构建链表
            else if (rl1->val<rl2->val) {
                p->next = new ListNode(rl1->val);
                rl1 = rl1->next;
                p = p->next;
            }
            else if (rl1->val >= rl2->val) {
                p->next = new ListNode(rl2->val);
                rl2 = rl2->next;
                p = p->next;
            }


        }
        return result->next;
    }
};