【LeetCode】【Python】【C++】2. Add Two Numbers代码实现

题目：

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

思路：

python中可以将两个链表转化为整型后进行相加，最后构造相加结果的单链表返回即可。

但是C++中因为数据溢出并不能这样实现。显然出题者意图并不是让我们按以上思路求解。

python实现：

# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        listnum1=[]
        listnum2=[]
        while l1 !=None:
            listnum1.append(l1.val)
            l1=l1.next

        while l2 !=None:
            listnum2.append(l2.val)
            l2=l2.next

        num1=0
        num2=0

        for i in range(len(listnum1)):
            num1=num1+listnum1[i]*(10**i)
        for i in range(len(listnum2)):
             num2 = num2 + listnum2[i] * (10 ** i)

        sum=num1+num2

        l3=ListNode(0)
        p = ListNode(0)
        p=l3

        while sum >=10:
            tem=ListNode(None)
            p.val=sum%10
            p.next=tem
            p=tem
            sum=sum//10

        p.val=sum
        return l3

C++实现：

#include
#include
using namespace std;


//Definition for singly-linked list.
struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
	__int64 trans(ListNode* l1)
	{
		__int64 a = 0;
		int i=0;
		while (l1 != NULL)
		{
			a = l1->val * pow(10,i)+a;
			l1 = l1->next;
			i++;
		}
		return a;
	}
	ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) 
	{
		__int64 a = trans(l1);
		__int64 b = trans(l2);
		__int64 c = a + b;
		ListNode* begin, *end, *p;
		begin = new ListNode(c % 10);
		end = begin;
		c = c / 10;
		while (c >= 1)
		{
			p = new ListNode(c%10);
			end->next = p;
			end = p;
			c = c / 10;	
		}	
		return begin;
	}
};

int main()
{
	Solution A;
	ListNode* l1, *l11, *l2, *l22,*p;
	l1 = new ListNode(1);
	l11 = l1;

	p = new ListNode(9);
	l11->next = p;
	l11 = p;

	p = new ListNode(9);
	l11->next = p;
	l11 = p;
	p = new ListNode(9);
	l11->next = p;
	l11 = p;

	p = new ListNode(9);
	l11->next = p;
	l11 = p;

	p = new ListNode(9);
	l11->next = p;
	l11 = p;

	p = new ListNode(9);
	l11->next = p;
	l11 = p;

	p = new ListNode(9);
	l11->next = p;
	l11 = p;

	p = new ListNode(9);
	l11->next = p;
	l11 = p;

	p = new ListNode(9);
	l11->next = p;
	l11 = p;




	l2 = new ListNode(9);
	l22 = l2;

	
	ListNode* result;
	result=A.addTwoNumbers(l1,l2);
	cout << "";
	return 0;
}

然而这样数据大的时候会出现溢出，于是下面是一种可接受的方法，思路是：对每一个链表元素分别相加，如有进位则设flag为1。

class Solution {
public:
	ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) 
	{
		ListNode*l3,*l33;
		l3 = new ListNode(0);
		l33 = l3;
		int flag = 0;
		while (l1!=NULL||l2!=NULL)
		{
			if (l1 == NULL)
			{
				ListNode* p;
				p = new ListNode((l2->val+flag)%10);
				l33->next = p;
				l33 = p;
				flag = (l2->val + flag) / 10;
				l2 = l2->next;
			}
			else if (l2 == NULL)
			{
				ListNode* p;
				p = new ListNode((l1->val+flag)%10);
				l33->next = p;
				l33 = p;
				flag = (l1->val + flag) / 10;
				l1 = l1->next;
			}
			else
			{
				ListNode* p;
				p = new ListNode((l1->val + l2->val + flag)%10);
				l33->next = p;
				l33 = p;
				flag = (l1->val + l2->val+flag) / 10;
				l1 = l1->next;
				l2 = l2->next;
			}

		}
		if (flag == 1)
		{
			ListNode* p;
			p = new ListNode(flag);
			l33->next = p;
			l33 = p;
		}
		l3 = l3->next;
		return l3;
		
	}
};