如何在 Java 中舍入整数除法并获得 int 结果?

How to round up integer division and have int result in Java?(如何在 Java 中舍入整数除法并获得 int 结果?)

问题描述

我刚刚写了一个小方法来计算手机短信的页数.我没有选择使用 Math.ceil 进行四舍五入，老实说，它看起来很丑.

I just wrote a tiny method to count the number of pages for cell phone SMS. I didn't have the option to round up using Math.ceil, and honestly it seems to be very ugly.

这是我的代码:

public class Main {

/**
 * @param args the command line arguments
 */
public static void main(String[] args) {
   String message = "today we stumbled upon a huge performance leak while optimizing a raycasting algorithm. Much to our surprise, the Math.floor() method took almost half of the calculation time: 3 floor operations took the same amount of time as one trilinear interpolation. Since we could not belive that the floor-method could produce such a enourmous overhead, we wrote a small test program that reproduce";

   System.out.printf("COunt is %d ",(int)messagePageCount(message));



}

public static double messagePageCount(String message){
    if(message.trim().isEmpty() || message.trim().length() == 0){
        return 0;
    } else{
        if(message.length() <= 160){
            return 1;
        } else {
            return Math.ceil((double)message.length()/153);
        }
    }
}

我不太喜欢这段代码，我正在寻找一种更优雅的方式来做这件事.有了这个，我期待 3 而不是 3.0000000.有什么想法吗?

I don't really like this piece of code and I'm looking for a more elegant way of doing this. With this, I'm expecting 3 and not 3.0000000. Any ideas?

推荐答案

你可以使用整数除法来四舍五入

To round up an integer division you can use

import static java.lang.Math.abs;

public static long roundUp(long num, long divisor) {
    int sign = (num > 0 ? 1 : -1) * (divisor > 0 ? 1 : -1);
    return sign * (abs(num) + abs(divisor) - 1) / abs(divisor);
}

或者如果两个数字都是正数

or if both numbers are positive

public static long roundUp(long num, long divisor) {
    return (num + divisor - 1) / divisor;
}

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