[leetcode] 2. Add Two Numbers

原创

是念 2022-08-11 17:55:41 博主文章分类：C++ ©著作权

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Description

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.

分析

首先创建一个head节点，不存储值，然后遍历l1和l2，然后根据加法运算法则来进行运算，注意里面很巧面的处理，把空的结点的值当作0来对待。

代码

/**
 * 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* dump=new ListNode(-1);
        ListNode* p=dump;
        int carry=0;
        while(l1||l2||carry!=0){
            int a=(l1==NULL ? 0:l1->val);
            int b=(l2==NULL ? 0:l2->val);
            int sum=a+b+carry;
            ListNode* node=new ListNode(sum%10);
            carry=sum/10;
            p->next=node;
            p=p->next;
            l1=(l1==NULL ? NULL:l1->next);
            l2=(l2==NULL ? NULL:l2->next);
        }
        return dump->next;
    }
};