https://oj.leetcode.com/problems/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
Keep track of the carry(进位) using a variable and simulate digits-by-digits sum from the head of list, which contains the least-significant digit(最低位).
Take extra caution of the following cases:
1. When one list is longer than the other.
2. The sum could have an extra carry of one at the end, which is easy to forget.(e.g., (9->9) + (1) = (0->0->1))
/** * Author : Acjx * Email : zhoujx0219@163.com */ /** * 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 *cur = dummyHead; // Present carry(进位) int carry = 0; // Process two lists and notice that one list may be longer than the other. 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; cur->next = new ListNode(digit % 10); cur = cur->next; if (p != NULL) p = p->next; if (q != NULL) q = q->next; } // The sum could have an extra carry of one at the end, which is easy to forget if (carry > 0) { cur->next = new ListNode(carry); } return dummyHead->next; } };
使用了 dummyHead，简化了代码。如果不使用 dummyHead，尾插需要判断头指针是否为空，示例代码如下：
ListNode *list = NULL; ListNode *tail = NULL; ... if (list == NULL) { // 记录新的链表头 list = new ListNode(sum); tail = list; } else { tail->next = new ListNode(sum); tail = tail->next; } ... return list;
显然，过于繁琐。
Using Dummy NodeScenario: When the head is not determinated.
原文：http://www.cnblogs.com/jianxinzhou/p/4312060.html