【Leetcode】【Easy】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.

解题：

题目比较简单，注意由于表头不确定，因此新建一个结点指向新创建的链表；

解题步骤：

1、判断l1和l2是否有效

2、新建preHead结点，newlist指针指向preHead；

3、循环开始，满足l1和l2都不为空：

　　（1）比较当前l1和l2的值，将较小的穿进newlist中

4、newlist链接l1或l2的剩余部分

5、释放表头，返回newlist；

/**
 * 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) {
        if (l1 == NULL) return l2;
        if (l2 == NULL) return l1;
        
        ListNode* prehead = new ListNode(0);
        ListNode* newlist = prehead;
        
        while (l1 != NULL && l2 != NULL) {
            if (l1->val > l2->val) {
                newlist->next = l2;
                l2 = l2->next;
            } else {
                newlist->next = l1;
                l1 = l1->next;
            }
            newlist = newlist->next;
        }
        
        if (l1 == NULL) newlist->next = l2;
        else newlist->next = l1;
        
        l1 = prehead->next;
        delete prehead;
        return l1;
    }
};