Java优化之输出十万以内的质数

(1)未经优化时所耗费的时间:

public class PrimeNumber {
    public static void main(String[] args) {
        long start = System.currentTimeMillis();
        boolean flag = false;
        for(int i = 2; i <= 100000; i++){
            for(int j = 2; j < i; j++){
                if(i % j == 0){
                    flag = true;
                }
            }
            if(flag == false)
                System.out.print(i+" ");
            flag = false;
        }
        long end = System.currentTimeMillis();
        System.out.println("\n"+(end - start));
    }
}

其所耗费的时间为:27038ms

(2)优化一:内层循环的判断,当false已经为true的时候,即可跳出内层循环

public class PrimeNumber {
    public static void main(String[] args) {
        long start = System.currentTimeMillis();
        boolean flag = false;
        for(int i = 2; i <= 100000; i++){
            for(int j = 2; j < i; j++){
                if(i % j == 0){
                    flag = true;
                    break;
                }
            }
            if(flag == false)
                System.out.print(i+" ");
            flag = false;
        }
        long end = System.currentTimeMillis();
        System.out.println("\n"+(end - start));
    }
}

 

其所耗费的时间为:2424ms

(3)优化二:内层循环只需循环到i的根号时,即可结束(注意包括i的根号)

public class PrimeNumber {
    public static void main(String[] args) {
        long start = System.currentTimeMillis();
        boolean flag = false;
        for(int i = 2; i <= 100000; i++){
            for(int j = 2; j <= Math.sqrt(i); j++){
                if(i % j == 0){
                    flag = true;
                    break;
                }
            }
            if(flag == false)
                System.out.print(i+" ");
            flag = false;
        }
        long end = System.currentTimeMillis();
        System.out.println("\n"+(end - start));
    }
}

 

其所耗费的时间为:191ms

 (4)也可以使用标签实现:

public class PrimeNumber2 {
    public static void main(String[] args) {
        long start = System.currentTimeMillis();
        label1:
        for(int i = 2; i <= 100000; i++){
            for(int j = 2; j <= Math.sqrt(i); j++){
                if(i % j == 0){
                    continue label1;
                }
            }
            System.out.print(i+" ");
        }
        long end = System.currentTimeMillis();
        System.out.println("\n"+(end - start));
    }
}

其耗费时间为:195ms

posted @ 2015-08-29 11:56  mengrennwpu  阅读(437)  评论(0编辑  收藏  举报