【刷題史】leetcode-Add Two Numbers

題目

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

code:

耗時：54ms

#include<iostream>
using namespace std;

struct ListNode
{
	int val;
	ListNode * next;
	ListNode(int x) :val(x), next(NULL){}
};
class Solution 
{
public:
	ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
	{
		ListNode * result = new ListNode(0);
		ListNode * p;
		ListNode * q = result;

		int tmp = 0;

		while (l1 != NULL||l2!= NULL)
		{
			if (l1 != NULL)
			{
				tmp += l1->val;
				l1 = l1->next;
			}
			if (l2 != NULL)
			{
				tmp += l2->val;
				l2 = l2->next;
			}
			p = new ListNode(0);
			if (tmp >= 10)
			{
				q->val = tmp % 10;
				tmp /= 10;
				p->val = tmp;
			}
			else
			{
				q->val = tmp;
				tmp = 0;
			}
			if (l1 != NULL || l2 != NULL||p->val!=0)
			{
				q->next = p;
				q = p;
			}
		}
		return result;
	}
};

class Solution1 {
public:
	ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
		// 題意可以認爲是實現高精度加法
		ListNode *head = new ListNode(0);
		//head->val = 0;
		//head->next = NULL;
		ListNode *ptr = head;
		int carry = 0;
		while (true) {
			if (l1 != NULL) {
				carry += l1->val;
				l1 = l1->next;
			}
			if (l2 != NULL) {
				carry += l2->val;
				l2 = l2->next;
			}
			ptr->val = carry % 10;
			carry /= 10;
			// 當兩個表非空或者仍有進位時需要繼續運算，否則退出循環
			if (l1 != NULL || l2 != NULL || carry != 0) {
				//ptr->next = new  ListNode;
				//ptr->next->val = 0;
				//ptr->next->next = NULL;
				ptr = (ptr->next = new ListNode(0));
			}
			else break;
		}
		return head;
	}
};

int main()
{
	ListNode * l1 = new ListNode(0);
	ListNode * l2 = new ListNode(0);
	ListNode * p = l1;
	ListNode *q = l2;
	ListNode *r;
	p->val = 1;
	//p = (p->next = new ListNode(4));
	//p = (p->next = new ListNode(3));
	q->val = 9;
	q = (q->next = new ListNode(9));
	//q = (q->next = new ListNode(4));
	//q = (q->next = new ListNode(6));
	Solution s;
	r=s.addTwoNumbers(l1, l2);
	int i = 0;
	while (r != NULL)
	{
		//cout << i << endl;
		cout << r->val << endl;;
		r = r->next;
	}

}