# LeetCode 2：两数相加 Add Two Numbers

2019.07.20 10:07 270浏览

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.

``````输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)

``````

### Java：

``````class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode head = new ListNode(0);//虚拟头节点
ListNode cur = head;//指针
int carry = 0;//进位值
while (l1 != null || l2 != null) {//两个链表均为空时停止遍历
int x = (l1 != null) ? l1.val : 0;//x为l1的值，如果节点为空，值为0
int y = (l2 != null) ? l2.val : 0;//y为l2的值，如果节点为空，值为0
int sum = carry + x + y;//sum为两节点值之和
carry = sum / 10;//得进位值（1）
cur.next = new ListNode(sum % 10);//sum%10 得余数即 个位数的值
cur = cur.next;//刷新指针
if (l1 != null) l1 = l1.next;//l1节点不为空继续刷新下一个节点
if (l2 != null) l2 = l2.next;//l2节点不为空继续刷新下一个节点
}
if (carry > 0) {//如果仍然需要进 1 ，则直接新建一个节点
cur.next = new ListNode(carry);
}
return head.next;
}
}
``````

### Python3：

``````class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
head = ListNode(0)
cur = head
sum = 0
while l1 or l2:
if l1:#l1不为空
sum += l1.val#累计两节点值的和
l1 = l1.next#刷新节点
if l2:
sum += l2.val#累计两节点值的和
l2 = l2.next#刷新节点
cur.next = ListNode(sum % 10)//刷新新链表
cur = cur.next
sum = sum // 10
if sum != 0:
cur.next = ListNode(sum)
return head.next
``````

0人点赞

• 推荐
• 评论
• 收藏
• 共同学习，写下你的评论

0/150