Discription
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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Solution
关键是处理好两个表的长度，以及进位问题，尤其是99+1=100这样特殊情况（和的长度比两个表的长度都大）的进位处理。
python
1 class Solution(object): 2 def addTwoNumbers(self, l1, l2): 3 """ 4 :type l1: ListNode 5 :type l2: ListNode 6 :rtype: ListNode 7 """ 8 testL1 = l1 9 testL2 = l2 10 carry = 0 11 root = current = ListNode(0) 12 13 while testL1 or testL2 or carry: 14 val1 = 0 15 val2 = 0 16 if testL1: 17 val1 = testL1.val 18 testL1 = testL1.next 19 20 if testL2: 21 val2 = testL2.val 22 testL2 = testL2.next 23 24 carry, mod_val = divmod(val1 + val2 + carry, 10) 25 current.next = ListNode(mod_val) 26 current = current.next 27 28 return root.next
cpp
1 class Solution { 2 public: 3 ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { 4 ListNode header(0); 5 ListNode *p = &header; 6 int carry = 0; 7 int sum = 0; 8 int mod = 0; 9 while (l1 || l2 || carry) 10 { 11 int val1 = 0; 12 int val2 = 0; 13 if (l1) 14 { 15 val1 = l1->val; 16 l1 = l1->next; 17 } 18 if (l2) 19 { 20 val2 = l2->val; 21 l2 = l2->next; 22 } 23 sum = val1 + val2 + carry; 24 carry = sum / 10; 25 mod = sum % 10; 26 p->next = new ListNode(mod); 27 p = p->next; 28 } 29 30 return header.next; 31 } 32 };