002 Add Two Numbers 链表上的两数相加

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.

详见：https://leetcode.com/problems/add-two-numbers/description/

Java实现：

/**

 * Definition for singly-linked list.

 * public class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode(int x) { val = x; }

 * }

 */

class Solution {

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        ListNode head=null;

        ListNode pre=null;

        int carry=0;

        while(l1!=null||l2!=null){

            int val1=l1!=null?l1.val:0;

            int val2=l2!=null?l2.val:0;

            int val=val1+val2+carry;

            carry=val/10;

            val%=10;

            ListNode cur=new ListNode(val);

            if(head==null){

                head=cur;

            }

            if(pre!=null){

                pre.next=cur;

            }

            pre=cur;

            l1=l1!=null?l1.next:null;

            l2=l2!=null?l2.next:null;

        }

        if(carry!=0){

            ListNode l=new ListNode(carry);

            pre.next=l;

        }

        return head;

    }

}

python：

# Definition for singly-linked list.

# class ListNode:

#     def __init__(self, x):

#         self.val = x

#         self.next = None

class Solution:

    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:

        head,pre,carry=None,None,0

        while l1 or l2:

            val1,val2=0,0

            if l1:

                val1=l1.val

                l1=l1.next

            if l2:

                val2=l2.val

                l2=l2.next

            val=val1+val2+carry

            carry=val//10

            val%=10

            cur=ListNode(val)

            if not head:

                head=cur

            if pre:

                pre.next=cur

            pre=cur

        if carry:

            pre.next=ListNode(carry)

        return head

巴特西

002 Add Two Numbers 链表上的两数相加

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