c++ - 两个非负整数A和B的十进制zip是整数C

Question

The decimal zip of two non-negative integers A and B is an integer C whose 
 decimal representation is created from the decimal representations
 of A and B as follows:

• the first (i.e. the most significant) digit of C is the first digit of A;
• the second digit of C is the first digit of B;
• the third digit of C is the second digit of A;
• the fourth digit of C is the second digit of B;
• etc.

 If one of the integers A and B runs out of digits, the remaining digits of 
 the other integer are appended to the result.

 The decimal representation of 0 is assumed to be "0".

 For example, the decimal zip of 12 and 56 is 1526.
 The decimal zip of 56 and 12 is 5162.
 The decimal zip of 12345 and 678 is 16273845.
 The decimal zip of 123 and 67890 is 16273890.

 Write a function: function solution(A, B); that, given two non-negative
 integers A and B, returns their decimal zip.

 The function should return -1 if the result exceeds 100,000,000.

 For example, given A = 12345 and B = 678 the function should return 
 16273845, as explained above.

假使，假设:

•A和B是[0..100,000,000]范围内的整数。

这是我的解决方案，但我得到了数组约束的期望

因此，我尝试将整数转换为字符串，以便对其进行处理，然后将数字加在一起。

int solution(int A, int B) {
    // write your code in C++11 (g++ 4.8.2)
    if (A < 0 || A > 100000000) return -1;
    if (B < 0 || B > 100000000) return -1;

    string A_ = IntToString(A);
    string B_ = IntToString(B);
    string output = "";

    for (int i = 0; i < A_.size() || i < B_.size(); i++) {

        if (A_[i]) {
            output[i] = output[i] + A_[i];
        }

        if (B_[i]) {
            output[i] = output[i] + B_[i];
        }
    }

    return atoi(output.c_str());
}

Answer 1

仅当for达到最大字符串的大小时，才会停止i循环。但是您继续使用i作为两个字符串的索引。因此，您将超出较短的范围，从而导致错误。

代替:

    if (A_[i])

您可能想要:

    if (i < A_.size())