2. Add Two Numbers 解题记录

题目描述：

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.

将两个链表里各节点的数相加，进位的进位，看例子就明白了。

解题思路：

这题的思路没有什么好说的，就是两个链表的数一个一个加起来。

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* head=new ListNode(0);
        ListNode* cur=head;
        int carry=0;
        while(l1!=NULL||l2!=NULL||carry){
        //l1,l1,进位都不存在时，出循环
            int num1=0,num2=0;
            if(l1){
            //l1存在是进行赋值，下同
                num1=l1->val;
                l1=l1->next;
            }
            if(l2){
                num2=l2->val;
                l2=l2->next;
            }
            num1+=num2+carry;//用num1存三个数的和
            cur->next=new ListNode(num1%10);
            carry=num1/10;
            cur=cur->next;
        }
        return head->next;
    }
};

 

他人代码：

/**
 * 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* head=new ListNode(0);
        ListNode* cur=head;
        int carry=0;
        while(l1!=NULL||l2!=NULL||carry){
            int n = (l1?l1->val:0) + (l2?l2->val:0) + carry;//运用了三目运算符使代码更简洁
            cur -> next = new ListNode(n % 10);
            carry = n / 10;
            cur = cur -> next;
            l1 = l1?l1 -> next:NULL;
            l2 = l2?l2 -> next:NULL;
        }
        return head->next;
    }
};