此文轉載自：https://blog.csdn.net/qq_36556893/article/details/110231369#commentBox

目錄

一、題目內容

二、解題思路

三、程式碼

一、題目內容

給你兩個 非空 連結串列來代表兩個非負整數。數字最高位位於連結串列開始位置。它們的每個節點只儲存一位數字。將這兩數相加會返回一個新的連結串列。

你可以假設除了數字 0 之外，這兩個數字都不會以零開頭。

進階：

如果輸入連結串列不能修改該如何處理？換句話說，你不能對列表中的節點進行翻轉。

示例：

輸入：(7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
輸出：7 -> 8 -> 0 -> 7

二、解題思路

哎，太難了，還是兩個函式來回套吧：

1.兩數相加：leetcode_2. 兩數相加

2.反轉連結串列：leetcode_206. 反轉連結串列

三、程式碼

# 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