lettcode21. Merge Two Sorted Lists

lettcode21. 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.

从小到大排列的两个数组，合并成一个数组。递归的方法。注意：两个数组都为空的情况。

This solution is not a tail-recursive, the stack will overflow while the list is too long

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

 2.新建了一个临时列表tmp，用时12ms，比上面多3ms（是什么原因呢？）

/**
 * 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(-1);
        ListNode *tmp=&head;
        while(l1&&l2){
            if(l1->val<l2->val){
                tmp->next=l1;
                l1=l1->next;
            }else{
                tmp->next=l2;
                l2=l2->next;
            }
            tmp=tmp->next;
        }
        if(l1)
            tmp->next=l1;
        if(l2)
            tmp->next=l2;
        return head.next;
    }
};