Leetcode合并两个有序链表

• 合并两个有序链表：将两个升序链表合并为一个新的升序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 

例如：

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

解法一：迭代法

/**
 * 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 *l3 = new ListNode;
        ListNode *s3 = l3;
        while(l1 != nullptr && l2 != nullptr){
            int min = 0;
            if(l1->val > l2->val){
                min = l2->val;
                l2 = l2->next;
            }
            else{
                min = l1->val;
                l1 = l1->next;
            }

            ListNode *newnode = new ListNode;
            newnode->val = min;
            l3->next = newnode;
            l3 = newnode;
        }

        if(l1 != nullptr){
            while(l1 != nullptr){
                ListNode *newnode = new ListNode;
                newnode->val = l1->val;
                l3->next = newnode;
                l3 = newnode;
                l1 = l1->next;
            }
        }

        if(l2 != nullptr){
            while(l2 != nullptr){
                ListNode *newnode = new ListNode;
                newnode->val = l2->val;
                l3->next = newnode;
                l3 = newnode;
                l2 = l2->next;
            }
        }

        return s3->next;
    }
};

解法二：递归

/**
 * 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 == nullptr) return l2;
        else if(l2 == nullptr) return l1;
        else if(l1->val < l2->val){
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        }
        else{
            l2->next = mergeTwoLists(l1, l2->next);
            return l2;
        }
    }
};