查找 2 个给定整数中位数之间差异的递归方法

Question

输入是两个给定的整数，都是N个数，输出必须是差的结果。例如，对于第一个数字是 1234，第二个数字是 1，输出必须是 3。我尝试编写一个递归方法，但它不让我减去它们，说需要两个整数但找到了一个。这是到目前为止的代码：

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

这是我的方法

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;
    }

输出示例：对于 a = 12345 和 b=1，输出将是：4

查找 2 个给定整数中位数之间差异的递归方法

Recursive method that finds the difference between the number of digits in 2 given ints

java

recursion

sum

digits