989. Add to Array-Form of Integer

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

这题有点像大数的四则运算，最简单的就是将容器转为整型然后直接计算，但是testcase里可能有很大的数，所以肯定是不符合题意的。不难，需要细心并且关注进位即可。

class AddToArrayFormOfInteger : public Solution {
public:
    void Execute() {
        auto res = addToArrayForm(vector<int>{9,9,9,9,9,9,9,9,9,9}, 1);
    }

    vector<int> addToArrayForm(vector<int> A, int K) {
        vector<int> res;

        bool adv = false;
        int i = A.size() - 1;
        while ((K != 0 || adv) && i >= 0) {
            int val = K % 10;

            int sum = A[i] + val + (adv ? 1 : 0);
            adv = false;
            A[i] = sum % 10;
            if (sum >= 10) {
                adv = true;
            }
            i--; K /= 10;
        }
        if (adv) {
            K++;
        }
        if (K != 0) {
            while (K != 0) {
                res.push_back(K % 10);
                K /= 10;
            }
            std::reverse(res.begin(), res.end());
            int psz = res.size();
            res.resize(res.size() + A.size());
            std::copy(A.begin(), A.end(), res.begin() + psz);
        } else {
            res = std::move(A);
        }
        return res;
    }
};