蜗牛慢慢爬 LeetCode 2. Add Two Numbers [Difficulty: Medium]

题目

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

翻译

给定两个非空的链表，表示两个非负整数。数字倒序存储，每个节点包含一个数字。将两个数字求和并将结果以链表形式返回
你可以假定数字不包含前导0（除了0本身之外）
输入：（2 - > 4 - > 3）+（5 - > 6 - > 4）
输出：7 - > 0 - > 8

Hints

Related Topics: Linked List, Math
注意简化代码

代码

Java

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode c1 = l1;
        ListNode c2 = l2;
        ListNode s = new ListNode(0);
        ListNode d = s;
        int sum = 0;
        while(c1!=null||c2!=null){
            sum /= 10;
            if(c1!=null){
                sum += c1.val;
                c1 = c1.next;
            }
            if(c2!=null){
                sum += c2.val;
                c2 = c2.next;
            }
            d.next = new ListNode(sum%10);
            d = d.next;
        }
        if(sum/10==1)
            d.next = new ListNode(1);
        return s.next;
    }
}

Python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        out = 0
        result = ListNode(0)
        l = result
        while l1!=None or l2!=None:
            x1 = l1.val if l1!=None else 0
            x2 = l2.val if l2!=None else 0
            l.next = ListNode((x1+x2+out)%10)
            out = (x1+x2+out)/10
            l = l.next
            
            l1 = l1.next if l1!=None else l1
            l2 = l2.next if l2!=None else l2
        if out!=0:
            l.next = ListNode(out)
        return result.next

//better solution from discuss
class Solution(object):        
    def addTwoNumbers(self, l1, l2):
        carry = 0
        root = n = ListNode(0)
        while l1 or l2 or carry:
            v1 = v2 = 0
            if l1:
                v1 = l1.val
                l1 = l1.next
            if l2:
                v2 = l2.val
                l2 = l2.next
            carry, val = divmod(v1+v2+carry, 10)
            n.next = ListNode(val)
            n = n.next
        return root.next