[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  : zhoujx0219@163.com
 */

/**
 * 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.