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LeetCode OJ - Add Two Numbers
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：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution {10 public:11 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {12 if (l1 == NULL) return l2;13 if (l2 == NULL) return l1;14 15 ListNode* head = new ListNode((l1->val + l2->val) % 10);16 int carry = (l1->val + l2->val) / 10;17 ListNode* p1 = l1->next, *p2 = l2->next;18 19 ListNode* cur = head;20 while (p1 != NULL && p2 != NULL) {21 int tmp = p1->val + p2->val + carry;22 cur->next = new ListNode(tmp % 10);23 carry = tmp / 10;24 p1 = p1->next;25 p2 = p2->next;26 cur = cur->next;27 }28 while (p1 != NULL) {29 int tmp = p1->val + 0 + carry;30 cur->next = new ListNode(tmp % 10);31 carry = tmp / 10;32 p1 = p1->next;33 cur = cur->next;34 }35 while (p2 != NULL) {36 int tmp = p2->val + 0 + carry;37 cur->next = new ListNode(tmp % 10);38 carry = tmp / 10;39 p2 = p2->next;40 cur = cur->next;41 }42 if (carry) {43 cur->next = new ListNode(carry);44 }45 return head;46 }47 };
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