Leetcode题解 - 2. Add Two Numbers

两数相加链表实现

最新推荐文章于 2020-01-30 21:46:13 发布

xuelians

最新推荐文章于 2020-01-30 21:46:13 发布

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CC 4.0 BY-SA版权

文章标签： leetcode C++

本文链接：https://blog.youkuaiyun.com/xuelians/article/details/54646243

本文介绍了一种解决两数相加问题的方法，通过链表形式存储逆序数字，并返回相加后的链表结果。该算法实现了进位计算，能够处理不同长度的输入链表。

题目

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

/**
 * 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) {
    // add code here
    }
}

本人的解

不算快，49 ms, Your runtime beats 9.95% of cpp submissions

/**
 * 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 *d1 = l1, *d2 = l2;
        ListNode *r = new ListNode(0), *p = r;

        do {
            int c = (d1 ? d1->val : 0) + (d2 ? d2->val : 0) + p->val;
            d1 = d1 ? d1->next : 0;
            d2 = d2 ? d2->next : 0;
            p->val = (c % 10);
            if( d1 || d2 || c >= 10 ) {
                p->next = new ListNode(c / 10);
                p = p->next;
            }
        } while( d1 || d2);
        return r;
    }
}