2. Add Two Numbers

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 contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zeros, except the number 0 itself.

Example 1:

1
2
3
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:

1
2
Input: l1 = [0], l2 = [0]
Output: [0]

Example 3:

1
2
Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

题目大意

给定两个逆序存储数字各位的非空链表（如 2->4->3 表示 342），将这两个数相加并以链表形式返回结果。假设两数除 0 外无前导零，需处理进位情况。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        cur = dummy = ListNode()  # 哨兵节点
        carry = 0  # 进位
        while l1 or l2 or carry:  # 有一个不是空节点，或者还有进位，就继续迭代
            if l1:
                carry += l1.val  # 节点值和进位加在一起
                l1 = l1.next  # 下一个节点
            if l2:
                carry += l2.val  # 节点值和进位加在一起
                l2 = l2.next  # 下一个节点
            cur.next = ListNode(carry % 10)  # 每个节点保存一个数位
            carry //= 10  # 新的进位
            cur = cur.next  # 下一个节点
        return dummy.next  # 哨兵节点的下一个节点就是头节点
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        dummy = ListNode(0)
        node = dummy  # node 一直会变化（前进）
        carrier = 0  # 进位

        while l1 or l2 or carrier:
            sum = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carrier

            node.next = ListNode(sum % 10)
            node = node.next
            
            carrier = sum // 10
            if l1: l1 = l1.next
            if l2: l2 = l2.next

        return dummy.next