c++ - 在 C++ 中使用递归方法反转整数

Question

我有这个递归函数来反转正整数。任何人有一个有效的算法来使用更少的递归调用(但仍在使用递归)来完成任务，请在这里发帖!

int countDigit (const int & _X) {
    if (_X < 10)
        return 1;
    return (countDigit(_X / 10) + 1);
}

int getPower (const int & _X, const int & _Y) {
    if (_Y == 0)
       return 1;
    int ans = getPower(_X, _Y / 2);
    if (_Y % 2)
       return _X * ans * ans;
    return ans * ans;
 }

 int reverseDigit (const int & digit)  {
    if (digit < 10) {
      return digit;
    }
    int num = (digit % 10) * getPower(10, countDigit(digit / 10));
    return num + reverseDigit(digit / 10);
 }

Answer 1

当然......有效地从号码中读取数字并将它们添加到临时号码的末尾:

EDIT: Based on the questioner's bizarre modification of the question (to require recursion), I have edited my clean iterative solution to include an arbitrary amount of recursion, denoted by the positive integer [recurse].

int reverseDigit(int digit, int recurse=1) {
    if (recurse>0) reverseDigit(digit,recurse-1);

    int num=0;
    while (digit!=0) {
        num*=10;
        num+=digit%10;
        digit/=10;
    }
    return num;
}