# Add to Array-Form of Integer Solution LeetCode

## Solution Add to Array-Form of Integer LeetCode

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`

SOLUTION:

``````auto speedup = []() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
return nullptr;
}();

class Solution {
public:
vector<int> addToArrayForm(vector<int>& A, int K) {
vector<int> ans;
int j = A.size() - 1, carry = 0;
while (K > 0 && j >= 0) {
int curr = A[j--] + carry + K % 10;
ans.push_back(curr % 10);
carry = curr / 10;
K /= 10;
}
while (j >= 0) {
int curr = A[j--] + carry;
ans.push_back(curr % 10);
carry = curr / 10;
}
while (K > 0) {
int curr = K % 10 + carry;
ans.push_back(curr % 10);
carry = curr / 10;
K /= 10;
}
if (carry > 0) ans.push_back(carry);
reverse(ans.begin(), ans.end());
return ans;
}
};``````