LeetCode 2. Add Two Numbers

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You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

题目很简单,把两个链表按元素加起来,注意进位,另外还要注意两个链表可能元素数量不等。

还有一个小陷阱就是最后可能会多一个元素。


解法一:

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode* root = new ListNode(0), *indexA = l1, *indexB = l2, *indexC = root;        int remain = 0;        while (indexA != NULL && indexB != NULL)        {            int total = indexA->val + indexB->val + remain;            indexC->val = total % 10;            remain = total / 10;            indexA = indexA->next;            indexB = indexB->next;            if (indexA != NULL || indexB != NULL)            {                indexC->next = new ListNode(0);                indexC = indexC->next;            }        }        while (indexA != NULL)        {            int total = indexA->val + remain;            indexC->val = total % 10;            remain = total / 10;            indexA = indexA->next;            if (indexA != NULL)            {                indexC->next = new ListNode(0);                indexC = indexC->next;            }        }        while (indexB != NULL)        {            int total = indexB->val + remain;            indexC->val = total % 10;            remain = total / 10;            indexB = indexB->next;            if (indexB != NULL)            {                indexC->next = new ListNode(0);                indexC = indexC->next;            }        }if (remain != 0)        {            indexC->next = new ListNode(remain);        }        return root;    }


解法过长而且用到了额外空间,想想应该可以不需要额外空间的。不过这个解法速度还不错,42ms超过了78.83%的人。



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