[LeetCode] 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
思路:两条链表相加,注意最后如果进位不为零,要加入到结果中。使用二级指针记录节点前缀。
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 || l2 == NULL) 13 return l1 == NULL ? l2 : l1; 14 15 int carry = 0; 16 ListNode *pter1 = l1; 17 ListNode *pter2 = l2; 18 ListNode *head = NULL; 19 ListNode **prev= &head; 20 while (pter1 != NULL || pter2 != NULL) { 21 const int l1_val = pter1 == NULL ? 0 : pter1->val; 22 const int l2_val = pter2 == NULL ? 0 : pter2->val; 23 int temp_val = l1_val + l2_val + carry; 24 ListNode *temp_node = new ListNode(temp_val % 10); 25 carry = temp_val / 10; 26 *prev = temp_node; 27 prev = &(temp_node->next); 28 pter1 = pter1 == NULL ? NULL : pter1->next; 29 pter2 = pter2 == NULL ? NULL : pter2->next; 30 } 31 32 if (carry == 1) { 33 ListNode *temp_node = new ListNode(1); 34 *prev = temp_node; 35 } 36 37 return head; 38 } 39 };