LeetCode 2: Add Two Numbers

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

解题思路

需要注意如下几点：
1. 数字是按逆序存放的；
2. 链表l1或l2可能为空；
3. 链表l1和l2长度可能不同；
4. 2个数相加，可能会产生最高位的进位，因此要注意在完成以上1－3的操作后，判断进位是否为0，不为0则需要增加结点存储最高位的进位。

/** * 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) {        if (l1 == NULL) return l2;        if (l2 == NULL) return l1;        int jinwei = 0;        ListNode *result = NULL, *tail = NULL;        while (l1 != NULL && l2 != NULL) {            int value = l1->val + l2->val + jinwei;            jinwei = value / 10;            ListNode *temp = new ListNode(value%10);            if (result == NULL) {                result = tail = temp;            }            else {                tail->next = temp;                tail =  temp;              }            l1 = l1->next;            l2 = l2->next;        }        while (l1 != NULL) {            int value = l1->val + jinwei;            jinwei = value / 10;            ListNode *temp = new ListNode(value%10);            tail->next = temp;            tail =  temp;            l1 = l1->next;        }        while (l2 != NULL) {            int value = l2->val + jinwei;            jinwei = value / 10;            ListNode *temp = new ListNode(value%10);            tail->next = temp;            tail =  temp;              l2 = l2->next;        }        if (jinwei > 0) {            ListNode *temp = new ListNode(jinwei);            tail->next = temp;            tail =  temp;          }        return result;    }};