Java program to display alphabets (A to Z) using loop

Count number of digits in an integer using while loop We can count the number of digits in a given number in many ways using java. One of the approach is that, we shall take the number, remove the last digit, and increment our counter for number of digits in the number. We can continue this, until there is no digit in the number. Another approach is that, we can convert the given number into a string, and use the method string.length() to count the number of character in this string. This results in the number of digits in given number. Here, while the loop is iterated until the test expression number != 0 is evaluated to 0(false). After the first iteration, number will be divided by 10 and its value will be 345. Then, the count is incremented to 1. After the second iteration, the value of number will be 34 and the count is increased to 2. After the third iteration, the value of number will be 3 and the count is increased to 3. After the fourth iteration, the value of number will be 0

Write a java program to check leap year Leap year contains 366 days, which comes once every four years. Every leap year corresponds to these facts: A century year is a year ending with 00.  A century year is a leap year only if it is divisible by 400. A leap year (except a century year) can be identified if it is exactly divisible by 4. A century year should be divisible by 4 and 100 both. A non-century year should be divisible only by 4. Let's find out whether a year is a leap year or not Program: public class Main { public static void main (String[] args) { // year to be checked int year = 2011 ; boolean leap = false ; // if the year is divided by 4 if (year % 4 == 0 ) { // if the year is century if (year % 100 == 0 ) { // if year is divided by 400 // then it is a leap year if (year % 400 == 0 ) leap = true ; else leap = false ; } // if the year is no