寻找List之和的最近素数

Posted 2020-12-26 ukzq

Task :

Given a List [] of n integers , find minimum mumber to be inserted in a list, so that sum of all elements of list should equal the closest prime number .

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Notes

• List size is at least 2 .

• List‘s numbers will only positives (n > 0) .

• Repeatition of numbers in the list could occur .

• The newer list‘s sum should equal the closest prime number .

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Input >> Output Examples

1- minimumNumber ({3,1,2}) ==> return (1)

Explanation:

• Since , the sum of the list‘s elements equal to (6) , the minimum number to be inserted to transform the sum to prime number is (1) , which will make the sum of the List equal the closest prime number (7) .

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2-  minimumNumber ({2,12,8,4,6}) ==> return (5)

Explanation:

• Since , the sum of the list‘s elements equal to (32) , the minimum number to be inserted to transform the sum to prime number is (5) , which will make the sum of the List equal the closest prime number (37) .

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3- minimumNumber ({50,39,49,6,17,28}) ==> return (2)

Explanation:

• Since , the sum of the list‘s elements equal to (189) , the minimum number to be inserted to transform the sum to prime number is (2) , which will make the sum of the List equal the closest prime number (191) .

 

public class Solution
{
    public static int minimumNumber(int[] numbers)
    {
        //首先求得现在list中的数值之和
        int sumOri = 0;
        
        for(int i=0;i<numbers.length;i++){
             sumOri += numbers[i];
        }
        //用另一个数对原始和进行递增
        int sumNow = sumOri;
        boolean searchPrime = true;
        //do...while递增原始数值,并判定是否为素数
        while(searchPrime)
        {
           
            int count = 0;
            for(int i=2;i<=Math.floor(sumNow/2);i++){
                if(sumNow%i==0){
                   //合数进入
                   count++;
                }
            }
           
            //count为0,这个数是质数,停止查找
            if(count == 0){
                 searchPrime = false;
            }else{
              //递增新的和
              sumNow++;
            } 
            
        }
        return sumNow-sumOri; // Do your magic!
    }
}