英文:
Advice on speeding up prime number counter
问题
以下是改进后的代码：
public static int countPrime(int limitNo) {
int noOfTimes = 0;
for (int o = 2; o <= limitNo; o++) {
boolean isPrime = true;
for (int i = 2; i <= Math.sqrt(o); i++) {
if (o % i == 0) {
isPrime = false;
break;
}
}
if (isPrime) {
noOfTimes++;
}
}
return noOfTimes;
}
英文:
How can I improve the logic of the following code? The purpose is to compute the number of primes from 1 to the user input (limitNo). The program works fine except it takes a while, more than the usual 1-3 secs, to generate a result for huge numbers like 99999.
public static int countPrime(int limitNo) {
int noOfTimes = 0, noOfRounds = 0; int o = 1;
while (o <= limitNo) {
for (int i = 1; i <= o; i++) {
if (o%i == 0) {
noOfRounds++;
}
}
if (noOfRounds == 2) {
noOfTimes++;
}
noOfRounds = 0;
o++;
}
return noOfTimes;
}
答案1
得分: 1
以下是翻译好的内容：
代码可以通过以下方式进行改进：
- 将部分代码行分离，以创建一个
isPrime()方法。 - 修改
for循环的限制条件，这样如果条件满足，意味着该数字不是质数。
public static boolean isPrime(int num) {
for (int i = 2; i < num; i++) {
if (num % i == 0) {
return false;
}
}
return true;
}
- 将方法中的代码替换为
isPrime()并更改起始值int o = 2;。
public static int countPrime(int limitNo) {
int noOfTimes = 0;
int o = 2;
while (o <= limitNo) {
if (isPrime(o)) {
noOfTimes++;
}
o++;
}
return noOfTimes;
}
- 当然，还有更好的改进方法，例如：
for (int i = 2; i <= num/2; i++)
for (int i = 2; i <= Math.sqrt(num); i++)
英文:
The code can be improved by
-
Separating some of the lines to make an
isPrime()method. -
Changing limits of the
forloop so that if the condition is met that means the number is not prime.public static boolean isPrime(int num) { for ( int i = 2 ; i < num ; i++ ) { if ( num % i == 0 ) { return false; } } return true; } -
Replacing the code in the method with
isPrime()and change the startint o = 2 ;.public static int countPrime(int limitNo) { int noOfTimes = 0; int o = 2; while ( o <= limitNo ) { if ( isPrime(o) ) { noOfTimes++; } o++; } return noOfTimes; } -
Of course, there are better and more improvements like:
for ( int i = 2 ; i <= num/2 ; i++ )for ( int i = 2 ; i <= Math.sqrt(num) ; i++ )
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