Question

我想知道在另一个号码中出现1个号码的次数。我找到了另一个号码中找到2位数字的解决方案，但我想要做的是找到提供的数字中的1位，2位，......，n位数字。我不想在开关指令中创建另一个案例，所以我的问题是如何避免切换以使其工作。代码如下：

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);

    System.out.println("Provide number:");
    int number1 = sc.nextInt();

    System.out.println("Provide number you want to find in number1:");
    int number2 = sc.nextInt();

    int counter = 0;
    int digit = 1;
    int a = 10;

    while(number2/a>0){
        digit++;
        a = a*10;
    }

    switch(digit){
        case 1:{
            while(number1 > 0 ){
                if(number1 % 10 == number2){
                    counter++;
                }
                number1 = number1/10;
            }
        }
        case 2:{
            while(number1 > 0){
                if(number1 % 100 == number2){
                    counter++;
                }
                number1 = number1/10;
            }
        }
    }
    System.out.println(counter);
}

感谢您的帮助。

Answer 1

嗯，你已经得到了答案。只需使用你的变量a。

while(number1 > 0){
   if(number1 % a == number2){
        counter++;
   }
   number1 = number1/10;
}

Answer 2

正如家里的工作，只是一些想法。

int entireNumber = 12345;
int soughtPartNumber = 234;

// Find the power of 10 that caps the soughtPartNumber (1000):
int cap = 1;
while (soughtPartNumber * 10 < cap) {
    cap *= 10;
}

// Search in the number as you did:
while (entireNumber >= soughtPartNumber) {
    ... (entireNumber % cap) == ...
    entireNumer /= 10; // Remove the right digit
}

替换该代码中的开关指令

2 个答案: