[leetcode 2] Add Two Numbers

最新推荐文章于 2022-02-22 21:07:37 发布

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#链表

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本文介绍了一种使用链表实现两个非负整数相加的方法。输入的两个链表以逆序方式存储数字，并且每个节点包含一个数字。文章提供了一个C++实现示例，展示了如何遍历两个链表，逐位相加并处理进位。

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

思路：用carry储存进位，不要漏掉最后一位，判断carry是0还是1

/**
 * 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) {
        if (!l1) {
            return l2;
        } else if (!l2) {
            return l1;
        }
        ListNode dummy(-1);
        int carry = 0;
        ListNode *prev = &dummy;
        for (auto p = l1, q = l2; p != NULL || q != NULL; p = p?p->next:NULL, q = q?q->next:NULL, prev = prev->next) {
            const int a = p?p->val:0;
            const int b = q?q->val:0;
            int value = (a+b+carry)%10;
            carry = (a+b+carry)/10;
            prev->next = new ListNode(value);
        }
        if (carry == 1) {
            prev->next = new ListNode(carry);
        }
        return dummy.next;
    }
};