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第七章第五题(打印不同的教)(Print distinct nu

第七章第五题(打印不同的教)(Print distinct nu

作者: in2013 | 来源:发表于2020-07-06 23:54 被阅读0次

    编写一个程序,读人10 个数并且显示互不相同的数(即一个数出现多次,但仅显示一次)。(提示,读人一个数,如果它是一个新数,则将它存储在数组中。如果该数已经在数组中,则忽略它。)输入之后,数组包含的都是不同的数。下面是这个程序的运行示例:

    Write a program that reads in ten numbers and displays
    the number of distinct numbers and the distinct numbers separated by exactly one
    space (i.e., if a number appears multiple times, it is displayed only once). (Hint:
    Read a number and store it to an array if it is new. If the number is already in the
    array, ignore it.) After the input, the array contains the distinct numbers.

    下面是参考答案代码:

    // https://cn.fankuiba.com
    import java.util.Arrays;
    import java.util.Scanner;
    
    public class Ans7_5_page236 {
        public static void main(String[] args) {
            Scanner input = new Scanner(System.in);
            System.out.print("Enter ten numbers: ");
            int[] numberList = new int[10];
            // int[] distinctList = {-1,-1,-1,-1,-1,-1,-1,-1,-1,-1};
            int[] distinctList = new int[10];
            Arrays.fill(distinctList,-1);
    
            for (int i = 0; i < 10; i++)
                numberList[i] = input.nextInt();
    
            for (int i = 0; i < 10; i++) {
                for (int j = 0; j < 10;j++) {
                    if (i == numberList[j])
                        distinctList[i] = numberList[j];
                }
            }
    
            int count = 0;
            for (int i = 0; i < 10; i++) {
                if (distinctList[i] != -1)
                    count++;
            }
            System.out.print("The number of distinct number is " + count+
                    "\nThe distinct numbers are: ");
    
            for (int i = 0; i < 10; i++) {
                if (distinctList[i] != -1) {
                    System.out.print(distinctList[i] + " ");
                }
            }
        }
    }
    

    适用Java语言程序设计与数据结构(基础篇)(原书第11版)Java语言程序设计(基础篇)(原书第10/11版)

    发布在博客:(https://cn.fankuiba.com)

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