本文介紹了如何在 Java 中舍入整數除法并獲得 int 結果?的處理方法,對大家解決問題具有一定的參考價值,需要的朋友們下面隨著小編來一起學習吧!
問題描述
限時送ChatGPT賬號..
我剛剛寫了一個小方法來計算手機短信的頁數.我沒有選擇使用 Math.ceil
進行四舍五入,老實說,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雅的方式來做這件事.有了這個,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);
}
或者如果兩個數字都是正數
or if both numbers are positive
public static long roundUp(long num, long divisor) {
return (num + divisor - 1) / divisor;
}
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