問題描述

我剛剛寫了一個(gè)小方法來計(jì)算手機(jī)短信的頁(yè)數(shù).我沒有選擇使用 Math.ceil 進(jìn)行四舍五入，老實(shí)說，它看起來很丑.

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.

這是我的代碼:

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);
        }
    }
}

我不太喜歡這段代碼，我正在尋找一種更優(yōu)雅的方式來做這件事.有了這個(gè)，我期待 3 而不是 3.0000000.有什么想法嗎?

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?

推薦答案

你可以使用整數(shù)除法來四舍五入

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);
}

或者如果兩個(gè)數(shù)字都是正數(shù)

or if both numbers are positive

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

這篇關(guān)于如何在 Java 中舍入整數(shù)除法并獲得 int 結(jié)果?的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！