[LeetCode] Add Two Numbers

https://oj.leetcode.com/problems/add-two-numbers/

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

Solution:

Keep track of the carry(进位) using a variable and simulate digits-by-digits sum from the head of list, which contains the least-significant digit(最低位).

Take extra caution of the following cases:

1. When one list is longer than the other.

2. The sum could have an extra carry of one at the end, which is easy to forget.(e.g., (9->9) + (1) = (0->0->1))

Code:

/**

 * Author : Acjx

 * Email  : [email protected]

 */



/**

 * Definition for singly-linked list.

 * struct ListNode {

 *     int val;

 *     ListNode *next;

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

 * };

 */

class Solution 

{

public:

    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) 

    {

        ListNode *dummyHead = new ListNode(0);

        

        ListNode *p = l1;

        ListNode *q = l2;

        ListNode *cur = dummyHead;

        

        // Present carry(进位)

        int carry = 0;

        

        // Process two lists and notice that one list may be longer than the other.

        while (p != NULL || q != NULL)

        {

            int x = (p != NULL) ? p->val : 0;

            int y = (q != NULL) ? q->val : 0;

            int digit = carry + x + y;

            

            carry = digit / 10;

            cur->next = new ListNode(digit % 10);

            cur = cur->next;

            

            if (p != NULL) p = p->next;

            if (q != NULL) q = q->next;

        }

        

        // The sum could have an extra carry of one at the end, which is easy to forget

        if (carry > 0)

        {

            cur->next = new ListNode(carry);

        }

        

        return dummyHead->next;

    }

};

Tips:

使用了 dummyHead，简化了代码。如果不使用 dummyHead，尾插需要判断头指针是否为空，示例代码如下：

    ListNode *list = NULL;

    ListNode *tail = NULL;

    ...

    if (list == NULL)

    {

        // 记录新的链表头

        list = new ListNode(sum);

        tail = list;

    }

    else

    {

        tail->next = new ListNode(sum);

        tail = tail->next;

    }

    ...

    return list;

显然，过于繁琐。

Using Dummy Node

Scenario: When the head is not determinated.