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题意:链表大数的加法处理,不过链表上的数是反序的。
本题来源:https://leetcode.com/problems/add-two-numbers/
分析:
1.如果其中链表为空,则不用计算了;如果两个链表都为空,则返回空(链表不熟)。
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 int carry = 0; 16 ListNode* tail = new ListNode(0); 17 ListNode* ptr = tail; 18 19 while(l1 != NULL || l2 != NULL){ 20 int val1 = 0; 21 if(l1 != NULL){ 22 val1 = l1->val; 23 l1 = l1->next; 24 } 25 26 int val2 = 0; 27 if(l2 != NULL){ 28 val2 = l2->val; 29 l2 = l2->next; 30 } 31 32 int tmp = val1 + val2 + carry; 33 ptr->next = new ListNode(tmp % 10); 34 carry = tmp / 10; 35 ptr = ptr->next; 36 } 37 if(carry == 1)ptr->next = new ListNode(1); 38 return tail->next; 39 } 40 };
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原文地址:http://www.cnblogs.com/orange1438/p/4377568.html