importjava.util.Scanner;publicclassMain{publicstaticvoidmain(String[]args){Scannerin=newScanner(System.in);System.out.print("Input an integer:");intn=in.nextInt();System.out.print("Check whether every digit of the said integer is even or not!\n");System.out.print(test(n));}publicstatic...
out.println("Provide a number:"); prdnum = oddevn.nextInt(); if (prdnum % 2 == 0) System.out.println("Provided number is even"); else System.out.println("Provided number is odd"); } } Output:This program will check whether a number is even or odd. The user provides the ...
publicclassOddNumbers{publicstaticvoidmain(String[]args){int[]numbers={1,2,3,4,5,6,7,8,9,10};for(intnumber:numbers){// 判断当前数字是否为偶数if(number%2==0){// 如果是偶数,则跳过本次循环continue;}// 打印奇数System.out.println(number);}}} 1. 2. 3. 4. 5. 6. 7. 8. 9. 1...
*/int c=ctl.get();/** * 1、如果当前线程数少于corePoolSize(可能是由于addWorker()操作已经包含对线程池状态的判断,如此处没加,而入workQueue前加了) */if(workerCountOf(c)<corePoolSize){//addWorker()成功,返回if(addWorker(command,true))return;/** ...
System.out.println("Prime check started for number: "+ crunchifyNumber); if(crunchifyNumber ==1){ return"1 is not a Prime number..."; } intlimit =(int)Math.sqrt(crunchifyNumber); for(inti =2; i<= limit; i++){ if(crunchifyNumber % i ==0){ ...
Check if String is Numeric with Regex Even though most developers will be content with using an already implemented method, sometimes you might have very specific pattern checking: publicstaticbooleanisNumeric(String string){// Checks if the provided string// is a numeric by applying a regular//...
int loop = 0; while (process.isAlive() && loop < 10) { loop++; Thread.sleep(1); System.out.println("loop: " + loop + ", still alive"); } System.out.println("进程信息" + process.isAlive() + process.exitValue()); System.out.println(stringBuilder.toString()); ...
So, what we need to do is to cast the result to int and multiply it by itself. Then, we check if the result is equal to the integer we started with: public static boolean isPerfectSquareByUsingSqrt(long n) { if (n <= 0) { return false; } double squareRoot = Math.sqrt(n); ...
if(n==1) { return true; } else { return false; } } Approach 2: We can use bitwise and operator to check if number is power of two or not. 1 2 3 4 5 6 public static boolean powerOfTwoBitwise(int n) { return (n & n-1)==0; } It will a very simple way to check if...
This even works for directories: Path tempDirectory = Files.createTempDirectory("baeldung-exists"); assertTrue(Files.exists(tempDirectory)); If we specifically want to know if a file or directory exists, we can also useFiles.isDirectory(Path)orFiles.isRegularFile(Path)methods: ...