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

time complexity O(n)
space complexity O(1)
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/**
 * 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 * pHead = new ListNode(0);
        ListNode * pi = pHead;
        int carry = 0;
        while( l1 != NULL || l2 != NULL){
            int sum = (l1==NULL?0:l1->val) + (l2==NULL?0:l2->val) + carry;
            carry = sum / 10;
            pi->next = new ListNode(sum%10);
            pi = pi->next;
            if( l1 != NULL)l1 = l1->next;
            if( l2 != NULL)l2 = l2->next;
        }
        if(carry == 1)
            pi->next = new ListNode(carry);
        return pHead->next;
    }
};





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原文地址：https://www.cnblogs.com/zhxshseu/p/e4251984041337fe10dcf7e242d90493.html