java - 基于整数的舍入问题

Question

Define the n-based integer rounding of an integer k to be the nearest multiple of n to k. If two multiples of n are equidistant use the greater one.

the 4-based rounding of 5 is 4 because 5 is closer to 4 than it is to 8,

the 5-based rounding of 5 is 5 because 5 is closer to 5 that it is to 10,

the 4-based rounding of 6 is 8 because 6 is equidistant from 4 and 8, so the greater one is used

Write a function named doIntegerBasedRounding that takes an integer array and rounds all its positive elements using n-based integer rounding. A negative element of the array is not modified and if n <=0, no elements of the array are modified. Finally you may assume that the array has at least two elements.

我在两个案例中面临的问题

({1, 2, 3, 4, 5},2) and ({-18, 1, 2, 3, 4, 5},4)

我得到了

[0, 2, 2, 4, 4] and [-18, 0, 0, 4, 4, 4]

代替

{2, 2, 4, 4, 6} and {-18, 0, 4, 4, 4, 4}

这是我的代码

class doIntegerBasedRounding {
public static void main(String[] args) {
    System.out.println(Arrays.toString(doIntegerBasedRounding(new int []{1,2,4,5,6,7,8,9,9},5)));
    System.out.println(Arrays.toString(doIntegerBasedRounding(new int []{1, 2, 3, 4, 5},2)));
    System.out.println(Arrays.toString(doIntegerBasedRounding(new int []{1, 2, 3, 4, 5},3)));
    System.out.println(Arrays.toString(doIntegerBasedRounding(new int []{1, 2, 3, 4, 5},-3)));
    System.out.println(Arrays.toString(doIntegerBasedRounding(new int []{-1, -2, -3, -4, -5},3)));
    System.out.println(Arrays.toString(doIntegerBasedRounding(new int []{-18, 1, 2, 3, 4, 5},4)));
    System.out.println(Arrays.toString(doIntegerBasedRounding(new int []{1, 2, 3, 4, 5},5)));
    System.out.println(Arrays.toString(doIntegerBasedRounding(new int []{1, 2, 3, 4, 5},100)));
    }

public static int[] doIntegerBasedRounding(int[]a, int n){
    int temp;
    int[] b= new int[a.length];
    if(n<0)
        return a;
    for (int i =0; i<a.length; i++){
        if(a[i]<=0){
            b[i]=a[i];
        }
        else if(a[i]>0)    {
        temp = a[i]%n;
            if(temp>n/2) {
                b[i] = a[i] + (n - temp);}
            if (temp<=n/2){
                b[i]= a[i]-temp;
            }
            }
        }
    return b;
    }
}

Answer 1

让我建议一个不同的方法:

调用你正在使用的值v:

1. 计算v除以n的余数
2. 从v中减去这个余数；这导致四舍五入
3. 如果余数等于或大于n/2，则添加n；这导致四舍五入。

在代码中，针对您的特定问题结构:

public void doIntegerBasedRounding(int[] values, int n)
{
    if (n > 0)
    {
        for (int i=0; i<values.length; i++)
        {
            if (values[i] > 0)
            {
                int rem = values[i]%n;
                values[i] = values[i] - rem + (rem>=n/2 ? n : 0);
            }
        }
    }
}