2. 两数相加

两数相加 #

题目 #

给出两个   非空 的链表用来表示两个非负的整数。其中,它们各自的位数是按照   逆序   的方式存储的,并且它们的每个节点只能存储   一位   数字。 如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。 您可以假设除了数字 0 之外,这两个数都不会以 0  开头。 示例: ``` 输入:(2 -> 4 -> 3) + (5 -> 6 -> 4) 输出:7 -> 0 -> 8 原因:342 + 465 = 807 ```
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. ```

分析 #


题解 #

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```python class Solution(object): def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ carry = 0 root = node = 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) node.next = ListNode(val) node = node.next return root.next ```
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```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: carry = 0 head = cur = ListNode(0) while l1 or l2: val1 = val2 = 0 if l1: val1 = l1.val l1 = l1.next if l2: val2 = l2.val l2 = l2.next tmp = val1 + val2 + carry carry = tmp // 10 cur.next = ListNode(tmp % 10) cur = cur.next if carry: cur.next = ListNode(carry) return head.next ```
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