1317. Convert Integer to the Sum of Two No-Zero Integers

Given an integer n. No-Zero integer is a positive integer which doesn't contain any 0 in its decimal representation.

Return a list of two integers [A, B] where:

A and B are No-Zero integers.
A + B = n
It's guarateed that there is at least one valid solution. If there are many valid solutions you can return any of them.

 

Example 1:

Input: n = 2
Output: [1,1]
Explanation: A = 1, B = 1. A + B = n and both A and B don't contain any 0 in their decimal representation.
Example 2:

Input: n = 11
Output: [2,9]
Example 3:

Input: n = 10000
Output: [1,9999]
Example 4:

Input: n = 69
Output: [1,68]
Example 5:

Input: n = 1010
Output: [11,999]
 

Constraints:

2 <= n <= 10^4

这题卡住了我一段时间。后来直接遍历每个数字了。我们只要求出一个减数就行了，保证减数和留下来的那位数字不为0。

数字小于等于1的，那么减数的位数我们置为数字+1,然后我们减去相应的值，然后再去遍历下一位的数字，循环往复。

假设到了最后一位，那么假设减数已经有值了，那么直接跳过生成即可。当然我觉得我这种方法算是笨办法，应当还有更加直接的方法，毕竟这题的难度是Easy。

class ConvertIntegerToTheSumOfTwoNoZeroIntegers : public Solution {
public:
    void Exec() {

    }
    vector<int> getNoZeroIntegers(int n) {
        vector<int> borrows;
        int num = n;
        while (num != 0) {
            int sub = num % 10;
            if (sub <= 1) {
                if (num / 10 == 0 && !borrows.empty()) {
                    break;
                }
                num -= sub + 1;
                borrows.push_back(sub + 1);
            } else {
                num -= 1;
                borrows.push_back(1);
            }
            num /= 10;
        }
        int fv = 0;
        for (int i = borrows.size() - 1; i >= 0; i--) {
            fv *= 10;
            fv += borrows[i];
        }
        return vector<int>{n - fv, fv};
    }
};