LeetCode OJ

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

解题思路：

直接一个个的加就行了，注意进位和边界处理。

代码：

/**
 * 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) {
        if (l1 == NULL) return l2;
        if (l2 == NULL) return l1;
        
        ListNode* head = new ListNode((l1->val + l2->val) % 10);
        int carry = (l1->val + l2->val) / 10;
        ListNode* p1 = l1->next, *p2 = l2->next;
        
        ListNode* cur = head;
        while (p1 != NULL && p2 != NULL) {
            int tmp = p1->val + p2->val + carry;
            cur->next = new ListNode(tmp % 10);
            carry = tmp / 10;
            p1 = p1->next;
            p2 = p2->next;
            cur = cur->next;
        }
        while (p1 != NULL) {
            int tmp = p1->val + 0 + carry;
            cur->next = new ListNode(tmp % 10);
            carry = tmp / 10;
            p1 = p1->next;
            cur = cur->next;
        }
        while (p2 != NULL) {
            int tmp = p2->val + 0 + carry;
            cur->next = new ListNode(tmp % 10);
            carry = tmp / 10;
            p2 = p2->next;
            cur = cur->next;
        }
        if (carry) {
            cur->next = new ListNode(carry);
        }
        return head;
    }
};