###### tags: `Array`, `Math`

# LeetCode 66.Plus One
You are given a large integer represented as an integer array ```digits```, where each ```digits[i]``` is the $i^{th}$ digit of the integer. The digits are ordered from most significant to least significant in left-to-right order. The large integer does not contain any leading ```0'```s.
Increment the large integer by one and return the resulting array of digits.

>Example 1:

Input: digits = [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Incrementing by one gives 123 + 1 = 124.
Thus, the result should be [1,2,4].

>Example 2:

Input: digits = [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
Incrementing by one gives 4321 + 1 = 4322.
Thus, the result should be [4,3,2,2].

>Example 3:

Input: digits = [9]
Output: [1,0]
Explanation: The array represents the integer 9.
Incrementing by one gives 9 + 1 = 10.
Thus, the result should be [1,0].

### Constraints:

- $1 <= digits.length <= 100$
- $-0 <= digits[i] <= 9$
- ```digits``` does not contain any leading ```0```'s.
---
### Idea:
>
### Solution:

Python:

class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        total = 0
        count = 0
        num = []
        for i in range(len(digits)-1,-1,-1):
            total += digits[i]*10**count
            count+=1
        tostr=str(total+1)
        for i in range(len(tostr)):
            num.append(tostr[i])
        return num

C++: