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 #








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
# 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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