LeetCode 2: Add Two Numbers

Description: 

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.

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描述：

 给定两个非空链表，分别表示两个非负整数。整数的组成数字按照逆序排列，并且链表的每个节点表示一个数字。将两个整数相加，并以链表的形式返回相加和。

假设两个整数均不包含前导0(除了整数0本身）。

例子：

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
解释： 342 + 465 = 807

遍历两个数字，并逐节点相加。以变量carrier表示进位。

/**
 * 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 *p1 = l1;
        ListNode *p2 = l2;
        int carrier = 0;
        ListNode *l3 = NULL;
        ListNode *p = NULL;
        
        while(p1 != NULL || p2 != NULL) {
            int val1 = p1 == NULL ? 0 : p1->val;
            int val2 = p2 == NULL ? 0 : p2->val;
            p1 = p1 == NULL ? NULL : p1->next;
            p2 = p2 == NULL ? NULL : p2->next;
            
            int val = val1 + val2 + carrier;
            carrier = val >= 10 ? 1 : 0;
            val = val >= 10 ? val - 10 : val;
            ListNode *node = new ListNode(val);
            if(l3 == NULL) {
                l3 = p = node;
            } else {
                p->next = node;
                p = node;
            }
        }
        if(carrier) {
            ListNode *node = new ListNode(carrier);
            p->next = node;
        }
        return l3;
    }
};

　　

时间复杂度为O(max(n1, n2)), 空间复杂度为O(max(n1, n2) + 1)。