21. 合并两个有序链表C++

迭代：新建一个头结点prehead。pre指针指向头节点prehead，然后将l1、l2中小的数加在pre指针后，直到l1、l2中有一个为空。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if(l1==NULL)return l2;
        if(l2==NULL)return l1;
        ListNode* prehead=new ListNode(-1);
        ListNode *pre;
        pre=prehead;
        while(l1!=NULL&&l2!=NULL)
        {
            if(l1->val>=l2->val)
            {
                pre->next=l2;
                l2=l2->next;
            }
            else
            {
                pre->next=l1;
                l1=l1->next;
            }
            pre=pre->next;
        }
        l1==NULL?pre->next=l2:pre->next=l1;
        return prehead->next;

    }
};