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 .
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 .
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) .
2- minimumNumber ({2,12,8,4,6}) ==> return (5)
- 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) .
3- minimumNumber ({50,39,49,6,17,28}) ==> return (2)
- 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!
}
}
将编程看作是一门艺术,而不单单是个技术。
敲打的英文字符是我的黑白琴键,
思维图纸画出的是我编写的五线谱。
当美妙的华章响起,现实通往二进制的大门即将被打开。