如何在Java中计算数组中的第一个重复值

Poonam Patil

我接受了数组int[] a = {33,33,5,5,9,8,9,9};在此数组中，有很多值重复，这意味着33次出现两次＆5次出现两次且9次出现3次。但我想计算重复的第一个值，即33是第一个出现两次的第一个值，因此答案应为2。

我尝试：

public class FindFirstDuplicate
{

public static void main(String[] args) {

    int c=0;
    int[] a = {33,33,5,5,9,8,9,9};

    outerloop:
    for(int i = 0; i < a.length; i++)
    {
        for(int j = i+1; j< a.length; j++)
        {
            if(a[i] == a[j])
            {
                System.out.println(a[i]); //Duplicate value
                c++;
                break outerloop;
            }
        }

    }
    System.out.print("Count: "+c);

    }
}

输出：33 1

狐狸

public class HelloWorld{
    public static void main(String[] args) {
        int[] a = {33,33,5,5,9,8,9,9};
        for(int i = 0; i < a.length; i++)
        {
            int c=1; // we already found one. 
                     // and we initialize this counter inside the loop, 
                     // so that it is reset for each new starting number.
            for(int j = i+1; j< a.length; j++) // we're starting from next number (reason we start with c=1)
            {
                if(a[i] == a[j])
                    c++;

            }
            if(c > 0) {
                System.out.println("First uplicate value: "+ a[i] + " Count: " + c);
                break;      // we have to break out of the outer loop, 
                            // so the inner loop can finish counting duplicates
            }
        }
    }
}

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