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) {
        int carry = 0;  
        ListNode* tail = new ListNode(0);  
        ListNode* ptr = tail;  
          
        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 tail->next;  
        
        
    }
};