2. Add Two Numbers

最新推荐文章于 2024-01-03 15:29:28 发布

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最新推荐文章于 2024-01-03 15:29:28 发布

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原文链接：http://www.cnblogs.com/Tycit/p/8032585.html

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的整数，并实现它们的相加运算。该算法适用于非负整数，通过遍历两个链表并逐位相加，同时处理进位，最终返回结果链表。

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.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

/**
 * 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 dummy(-1);
        ListNode *head = &dummy;
                    
        int carry = 0;
        while( l1 || l2 ){
            const int n1 = l1?l1->val:0;
            const int n2 = l2?l2->val:0;       
            const int value = (n1 + n2 + carry) % 10;
            carry = (n1 + n2 + carry) / 10; // 进位是下次计算算上 故要写在value下
            
            head->next = new ListNode(value);
            
         //   l1 = l1->next?l1->next:nullptr; //本身是否为空指针没判断 会产生空指针
         //  l2 = l2->next?l2->next:nullptr;
            
            l1 = l1?l1->next:nullptr;
            l2 = l2?l2->next:nullptr;
            
            head = head->next;
        }
        if(carry == 1)
            head ->next = new ListNode(1);
        return dummy.next;
    }
};