标签：
题目描述：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
注意事项：
1. (2 -> 4 -> 3)中，2是最低位，3是最高位；
2. 两个list的长度不一样；
3. 进位，(9 -> 9) + (1) = (0 -> 0 -> 1)；
4. 不用考虑负数的情况。
结题思路：
1. 定义一个ListNode用来存放结果（result）；
2. 相同位相加，将计算结果的个位传给result，进位传给carry。指针分别后移，如果不为NULL则循环；
3. 最后再次判断进位carry；
4. 返回结果。
代码如下（参考网上）：
/** * 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) { int carry = 0; //进位标志 ListNode* result = new ListNode(0); ListNode* ptr = result; while(l1 != NULL || l2 != NULL){ int val1 = 0; if(l1 != NULL){ val1 = l1->val; l1 = l1->next; } int val2 = 0; if(l2 != NULL){ val2 = l2->val; l2 = l2->next; } int tmp = val1 + val2 + carry; ptr->next = new ListNode(tmp % 10); carry = tmp / 10; ptr = ptr->next; } if(carry == 1){ ptr->next = new ListNode(1); } return result->next; } };
/** * 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) { ListNode* dummyHead = new ListNode(0); ListNode* p = l1; ListNode* q = l2; ListNode* curr = dummyHead; int carry = 0; while(p != NULL || q != NULL){ int x = (p != NULL) ? p->val : 0; int y = (q != NULL) ? q->val : 0; int digit = carry + x + y; carry = digit / 10; curr->next = new ListNode(digit % 10); curr = curr->next; if(p != NULL) p = p->next; if(q != NULL) q = q->next; } if(carry > 0) curr->next = new ListNode(carry); return dummyHead->next; } };
标签：
原文地址：http://www.cnblogs.com/510602159-Yano/p/4243412.html
