【LeetCode】2. 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】

利用两个指针分别遍历两个链表，并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可，若最后有进位需要添加一位。

/** * 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 && l2 == NULL) return NULL;        ListNode* head = new ListNode(0);        ListNode* tmp = head;        int add = 0;        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 sum = val1 + val2 + add;            tmp->next = new ListNode(sum % 10);            if(sum>=10)                add=1;            else add = 0;            tmp =tmp->next;                   }        if(add==1)            tmp->next = new ListNode(1);        return head->next;    }};