Java程序来检查一个数字是否是霓虹数字

霓虹数字是数字的平方的位数之和等于该数字的数字。任务是检查和打印范围内的霓虹灯数字。

插图：

算法 ：

1. 首先，找到给定数字的平方。
2. 使用循环求平方数的总和。
   
3. 条件校验和等于给定的数字
   1. 返回真
   2. 否则返回假。

Case 1:

Input  : 9
Output : Given number  9 is Neon number

Explanation : square of 9=9*9=81;
              sum of digit of square : 8+1=9(which is equal to given number)


Case 2:

Input : 8
Output : Given number is not a Neon number
 
Explanation : square of 8=8*8=64
              sum of digit of square : 6+4=10(which is not equal to given number)

例子：

Java

Pseudo code : Square =n*n;
              while(square>0)
              {
              int r=square%10;
              sum+=r;
              square=square/10;
              }

输出

// Java Program to Check If a Number is Neon number or not
 
// Importing java input/output library
import java.io.*;
 
class GFG {
 
    // Method to check whether number is neon or not
    // Boolean type
    public static boolean checkNeon(int n)
    {
        // squarring the number to be checked
        int square = n * n;
 
        // Initializing current sum to 0
        int sum = 0;
 
        // If product is positive
        while (square > 0) {
 
            // Step 1: Find remainder
            int r = square % 10;
 
            // Add remainder to the current sum
            sum += r;
 
            // Drop last digit of the product
            // and store the number
            square = square / 10;
        }
 
        // Condition check
        // Sum of digits of number obtained is
        // equal to original number
        if (sum == n)
 
            // number is neon
            return true;
        else
 
            // numbe is not neon
            return false;
    }
 
    // Main driver method
    public static void main(String[] args)
    {
        // Custom input
        int n = 9;
 
        // Calling above function to check custom number or
        // if user entered number via Scanner class
        if (checkNeon(n))
 
            // Print number considered is neon
            System.out.println("Given number " + n
                               + " is Neon number");
        else
 
            // Print number considered is not neon
            System.out.println("Given number " + n
                               + " is not a Neon number");
    }
}

时间复杂度： O(l) 其中 l 是给定数字的平方中的数字