leetcode_445. 两数相加 II

此文转载自：https://blog.csdn.net/qq_36556893/article/details/110231369#commentBox

目录

一、题目内容

二、解题思路

三、代码

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一、题目内容

给你两个 非空 链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头。

进阶：

如果输入链表不能修改该如何处理？换句话说，你不能对列表中的节点进行翻转。

示例：

输入：(7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 8 -> 0 -> 7

二、解题思路

哎，太难了，还是两个函数来回套吧：

1.两数相加：leetcode_2. 两数相加

2.反转链表：leetcode_206. 反转链表

三、代码

# 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:
        head = self._addTwoNumbers(self.reverseList(l1), self.reverseList(l2))
        return self.reverseList(head)

    def _addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        head = ListNode(0)
        # print(head.val, head.next)
        p = head
        q = p
        carry = 0
        while l1 or l2:

            if l1 is not None:
                l1_val = l1.val
                # print("l1:", l1_val)
            else:
                l1_val = 0
            if l2 is not None:
                l2_val = l2.val
                # print("l2:", l2_val)
            else:
                l2_val = 0
            temp = l1_val + l2_val + carry
            if temp >= 10:
                carry = 1
                p.val = temp - 10
                p.next = ListNode(1)
            else:
                carry = 0
                p.val = temp
                p.next = ListNode(0)
            q = p
            p = p.next
            if l1 is not None:
                l1 = l1.next
            if l2 is not None:
                l2 = l2.next
        if q.next.val == 0:
            q.next = None
            del p
        return head

    def reverseList(self, head: ListNode):
        # null     1 --> 2 --> 3 --> 4 --> null
        # null <-- 1 <-- 2 <-- 3 <-- 4     null
        if not head:
            return None
        cur, cur_new_next = head, None
        while cur:
            # save original next node
            origin_next = cur.next
            # link new next node
            cur.next = cur_new_next
            # current node turns to new next node
            cur_new_next = cur
            # original next node turns to current node
            cur = origin_next
        return cur_new_next