刷题21. Merge Two Sorted Lists

一、题目说明

这个题目是21. Merge Two Sorted Lists，归并2个已排序的列表。难度是Easy！

二、我的解答

既然是简单的题目，应该一次搞定。确实1次就搞定了，但是性能太差：

Runtime: 20 ms, faster than 8.74% of C++ online submissions for Merge Two Sorted Lists.
Memory Usage: 9.4 MB, less than 5.74% of C++ online submissions for Merge Two Sorted Lists.

代码如下：

class Solution{
	public:
		ListNode* mergeTwoLists(ListNode* l1, ListNode* l2){
			if(l1 ==NULL && l2==NULL) return NULL;
			if(l1 !=NULL && l2==NULL) return l1;
			if(l1 ==NULL && l2!=NULL) return l2;
			
			ListNode dummy(-1);
			ListNode* p = &dummy;
			while(l1 !=NULL && l2 !=NULL){
				if(l1->val <= l2->val){
				     p->next = l1;
				     p = p->next;
				     l1 = l1->next;
				}else{
					p->next = l2;
					p = p->next;
				    l2 = l2->next;
				}
			}
			
			if(l1 !=NULL){
			    p->next = l1;
			}
			
			if(l2 !=NULL){
			    p->next = l2;
			}
			
			return dummy.next;
		}
};

三、优化措施

优化后，8s，代码如下：

#include<iostream>
using namespace std;
struct ListNode{
	int val;
	ListNode*next;
	ListNode(int x):val(x),next(NULL){
	}
};

/**
 * 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(NULL == l1) return l2;
            if(NULL == l2) return l1;
			
			ListNode dummy(-1);
			ListNode* p = &dummy;
			while(l1  && l2 ){
				if(l1->val <= l2->val){
				     p->next = l1;
				     l1 = l1->next;
				}else{
					p->next = l2;
				    l2 = l2->next;
				}
                p = p->next;
			}
			
			p->next = l1 ? l1 : l2;
			
			return dummy.next;
		}
};

int main(){
	Solution s;
	ListNode* l1,*l2;
	ListNode* tmp;
	
	//init l1
	tmp = new ListNode(4);
	l1 = tmp;
	
	tmp = new ListNode(2);
	tmp->next = l1;
	l1 = tmp;
	
	tmp = new ListNode(1);
	tmp->next = l1;
	l1 = tmp;
	
	//init l2
	tmp = new ListNode(4);
	l2 = tmp;
	
	tmp = new ListNode(3);
	tmp->next = l2;
	l2 = tmp;
	
	tmp = new ListNode(1);
	tmp->next = l2;
	l2 = tmp;
	
	
	ListNode* l3 = s.mergeTwoLists(l1,l2);
	while(l3!=NULL){
		cout<<l3->val<<" ";
		l3 = l3->next;
	}
	return 0;
}