找到2個給定整數中位數之差的遞歸方法

Question

輸入是兩個給定的整數，都是 N 個數字，output 必須是差異的結果。 例如，對於第一個數字是 1234，第二個是 1，output 必須是 3。我嘗試編寫遞歸方法，但它不會讓我減去它們，說需要兩個整數，但找到了一個。 這是到目前為止的代碼：

public static int digitDiffer (int a, int b){
   int sumA = 0;
   int sumB = 0;
   if(a == 0 && b==0){
     return 0;
   }
   else {
     sumA += a % 10;
     a /= 10;
     sumB += b % 10;
     b /= 10;
   }
   return digitDiffer (sumA-sumB);
 }

Answer 1

這是我的方法

public static int digitDiffer (int a, int b){
        // I will divide a and b by 10 until one of them is 0
        if(a != 0 && b != 0) return digitDiffer(a/10, b/10);

        //if b == 0 and a != 0 then I will count how many times I can divide a by 10 until it becomes 0
        if(a != 0 && b == 0) return 1 + digitDiffer(a/10, 0);

        // if a == 0 and b != 0 then I will count how many times I can divide b by 10 until it becomes 0
        if(a == 0 && b != 0) return 1 + digitDiffer(0, b/10);
        return 0;
    }

Output 示例：對於a = 12345和b=1 ，output 將為： 4