Description

For a non-negative integer X, the array-form of X is an array of its digits in left to right order. For example, if X = 1231, then the array form is [1,2,3,1].

Given the array-form A of a non-negative integer X, return the array-form of the integer X+K.

Example 1:

Input: A = [1,2,0,0], K = 34
Output: [1,2,3,4]
Explanation: 1200 + 34 = 1234

Example 2:

Input: A = [2,7,4], K = 181
Output: [4,5,5]
Explanation: 274 + 181 = 455

Example 3:

Input: A = [2,1,5], K = 806
Output: [1,0,2,1]
Explanation: 215 + 806 = 1021

Example 4:

Input: A = [9,9,9,9,9,9,9,9,9,9], K = 1
Output: [1,0,0,0,0,0,0,0,0,0,0]
Explanation: 9999999999 + 1 = 10000000000

Note:

  1. 1 <= A.length <= 10000.
  2. 0 <= A[i] <= 9.
  3. 0 <= K <= 10000.
  4. If A.length > 1, then A[0] != 0

分析

题目的意思是：给定一个数组，和一个数，返回其相加后的结果，这是一个模拟加法的题目，我以前碰见过，于是就自己实现了一下，时间复杂度为O(n)，应该没有比这个更好的算法了，代码写得不够简洁，看了一下标准答案，用到了divmod函数来求余数和商，另外，它并没有拆分K，而是用K一步一步的跟A的数相加，最后得到的结果，真是精妙啊,思路清楚了以后我感觉我也能写得出来，所以就没有按照参考答案来了。

代码

class Solution:
    def addToArrayForm(self, A: List[int], K: int) -> List[int]:
        arr=[]
        if(K==0):
            arr.append(0)
        while(K>0):
            arr.append(K%10)
            K=K//10
        m=len(A)
        A.reverse()
        n=len(arr)
        min_len=min(m,n)
        res=[]
        t=0
        i=0
        for i in range(min_len):
            t+=arr[i]+A[i]
            res.append(t%10)
            t=t//10
        i+=1
        while(i<m):
            t+=A[i]
            i+=1
            res.append(t%10)
            t=t//10
        while(i<n):
            t+=arr[i]
            i+=1
            res.append(t%10)
            t=t//10
        if(t>0):
            res.append(t)
        res.reverse()
        return res

参考文献

​​[LeetCode] Approach 1: Schoolbook Addition​​