题目

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

/**
 * 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) {

    }
};

思路

• 只使用O(1)的额外空间，将相加的和存到l1中，返回l1链表头。

代码

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* head{l1};
        int carry{0};
        int value{0};
        while(l1->next != nullptr && l2->next != nullptr)
        {
            value = l1->val + l2->val + carry;
            carry = value / 10;
            value %= 10;
            l1->val = value;
            l1 = l1->next;
            l2 = l2->next;
        }
        //碰到l1或者l2的最后一个节点
        value = l1->val + l2->val + carry;
        carry = value / 10;
        value %= 10;
        l1->val = value;
        //如果l2的后继非空，则修改l1->next指向l2的后继
        if ((l2 = l2->next) != nullptr)
        {
            l1->next = l2;
        }
        while (carry && l1->next)
        {
        //当carry非0 并且l1的后继非空，则转移l1至它的后继
            l1 = l1->next;
            value = l1->val + carry;
            carry = value / 10;
            value %= 10;
            l1->val = value;
        }
        if (carry)
        {
            l1->next = new ListNode(carry);
        }
        return head;
    }
};

运行时间约30ms，不需要创建新节点，没有多余的代码，不知道为什么这么慢。。