一.题目

Add Two Numbers

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Submissions: 243190My
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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

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二.解题技巧

这道题是一个简单的链表遍历的问题，然后将不同将对应位置的两个数字进行相加，同时考虑进位的问题，并没有多少的算法在里面，只不过是为了遍历的方便，加了一个临时的链表头部节点，用于方便将第一个节点的遍历也并入到其他节点的遍历中，同时，考虑到两个链表的长度不一样的情况。

三.实现代码

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

#include <iostream>

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution
{
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
    {
        // return null if the list is null.
        if (l1 == NULL || l2 == NULL)
        {
            return NULL;
        }

        // the start node
        ListNode* ResultList = new ListNode(10);

        int Carry = 0;
        ListNode* Tail = ResultList;
        //ListNode* Tail = l1;
        int Result = 0;

        while ((l1 != NULL) && (l2 != NULL))
        {
            Result = l1->val + l2->val + Carry;
            Carry = Result / 10;
            Result = Result % 10;

            ListNode* Tmp = new ListNode(Result);
            Tail->next = Tmp;
            Tail = Tmp;
            l1 = l1->next;
            l2 = l2->next;
        }

        while (l1 != NULL)
        {
            Result = l1->val + Carry;
            Carry = Result / 10;
            Result = Result % 10;

            ListNode* Tmp = new ListNode(Result);
            Tail->next = Tmp;
            Tail = Tmp;
            l1 = l1->next;
        }

        while (l2 != NULL)
        {
            Result = l2->val + Carry;
            Carry = Result / 10;
            Result = Result % 10;

            ListNode* Tmp = new ListNode(Result);
            Tail->next = Tmp;
            Tail = Tmp;
            l2 = l2->next;
        }

        // add the carry
        if (Carry == 1)
        {
            ListNode* Tmp = new ListNode(1);
            Tail->next = Tmp;
        }

        //remove the start node
        ListNode* Tmp = ResultList->next;
        delete ResultList;
        ResultList = Tmp;

        return ResultList;

    }
};

void Test()
{
    ListNode *l1 = new ListNode(3);
    ListNode *l2 = new ListNode(5);

    Solution S1;
    ListNode *result = S1.addTwoNumbers(l1, l2);

    std::cout << "The result is " << result->val << std::endl;

}

四.体会

这道题对于算法并没有很高的要求，这是一道比较简单的大数相加和链表遍历的题，只要注意更新指向当前节点的指针，避免出现死循环就可以了。

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