1. 题目

leetcode.cn/problems/ad…

2. 解析

字符串相加，采用末尾相加，直到加到两个数的第一位，最后需要判断商是否大于0。使用双指针解法

3. 核心代码

class Solution(object):
    def add(self, a, b):
        al = len(a) - 1
        bl = len(b) - 1
        sha = 0
        result = ''
        while al >= 0 or bl >= 0:
            if al >= 0:
                i = int(a[al])
            else:
                i = 0
            if bl >= 0:
                j = int(b[bl])
            else:
                j = 0
            num = i + j + sha
            y = num % 10
            sha = num // 10
            result += str(y)
            al -= 1
            bl -= 1
        if sha != 0:
            result += str(sha)
        return result[::-1]


if __name__ == '__main__':
    s = Solution()
    print(s.add('9', '1'))

4. 题目变形

leetcode.cn/problems/ad…

5. 解析

单链表的加法，利用双指针的形式进行每个位数的遍历，位数需要一一对应相加，链表尝尝需要使用一个0开始的头最后返回result.next作为结果

6. 核心代码

# Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next


from typing import Optional


class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        result = cur = ListNode()
        sha = 0
        while l1 or l2:
            i = l1.val if l1 else 0
            j = l2.val if l2 else 0
            num = i + j + sha
            sha = num // 10
            r = num % 10
            cur.next = ListNode(r)
            if l1: l1 = l1.next
            if l2: l2 = l2.next
            cur = cur.next
        if sha: cur.next = ListNode(sha)
        return result.next


if __name__ == '__main__':
    s = Solution()
    x = ListNode(9, ListNode(9, ListNode(9, ListNode(9))))
    y = ListNode(9, ListNode(9, ListNode(9)))
    xx = s.addTwoNumbers(x, y)
    while xx:
        print(xx.val)
        xx = xx.next
    print(9999 + 999)