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Java Exercises: Check if a string starts with a specified word

Java Basic: Exercise-85 with Solution

Write a Java program to check if a string starts with a specified word.

Sample Data: string1 = "Hello how are you?"

Pictorial Presentation:

Java Basic Exercises: Check if a string starts with a specified word

Sample Solution:

Java Code:

import java.util.*; 
import java.io.*; 
 public class Exercise85 {
 public static void main(String[] args)
 {
   String string1 = "Hello how are you?";
    System.out.println(string1.startsWith("Hello"));
  }
}

Sample Output:

true

Flowchart:

Flowchart: Java exercises: Check if a string starts with a specified word

Java Code Editor:

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Previous: Write a Java program to take the last three characters from a given string and add the three characters at both the front and back of the string.
Next: Write a Java program start with an integer n, divide n by 2 if n is even or multiply by 3 and add 1 if n is odd, repeat the process until n = 1.

What is the difficulty level of this exercise?



Java: Tips of the Day

How to detect a loop in a linked list?

You can make use of Floyd's cycle-finding algorithm, also known as tortoise and hare algorithm.

The idea is to have two references to the list and move them at different speeds. Move one forward by 1 node and the other by 2 nodes.

  • If the linked list has a loop they will definitely meet.
  • Else either of the two references(or their next) will become null.

Java function implementing the algorithm:

boolean hasLoop(Node first) {

    if(first == null) // list does not exist..so no loop either
        return false;

    Node slow, fast; // create two references.

    slow = fast = first; // make both refer to the start of the list

    while(true) {

        slow = slow.next;          // 1 hop

        if(fast.next != null)
            fast = fast.next.next; // 2 hops
        else
            return false;          // next node null => no loop

        if(slow == null || fast == null) // if either hits null..no loop
            return false;

        if(slow == fast) // if the two ever meet...we must have a loop
            return true;
    }
}

Ref: https://bit.ly/2P6SCq4