java - 如果输入数字已经在数组 "Retry"中，则将不同的值数字放入数组中

Question

以下代码仅当连续输入的数字相同(例如1, 1或2, 2)时才可以请求重试。但是如果输入是例如1, 2, 1或2, 1，则不会请求重试, 3, 2。

import java.util.Scanner;
public class exercise {

    public static void main(String[] args) {
        Scanner scan = new Scanner (System.in);
        int[] a = new int[5];
        boolean found = false;
        int i = 1;
        System.out.println("Enter Number.\n>>>");
        a[0] = scan.nextInt();
        while(i<5)
        {
            System.out.println("Enter number.\n>>>");
            int num = scan.nextInt();
            for (int j=0; j<i; j++)  // to determine same number is already in array or not
            {
                if(a[j] == num)
                {
                    found = false;
                }
                else
                {
                    found = true;
                }
            }
            if (!found)
            {
                System.out.println("Retry");
            }
            else
            {
                a[i] = num;
                i++;
            }
        }
        for (int a1 : a)
            System.out.println(a1);
    }

}

Answer 1

找到数组中的元素后，中断循环。即使在数组中找到元素后，您仍会进行迭代。你的内循环应该是这样的:

  for (int j=0; j<i; j++) {
    if(a[j] == num) {
      found = false;
      break;          // break once you find the element.
    }
    else {
      found = true;
    }
  }