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  • 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
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