Loops are often used for String Traversals or String Processing where the code steps through a string character by character. In previous sections we saw how to use String objects and built-in string methods to process strings. In this lesson, we will write our own loops to process strings.
String substring(int from, int to) : returns the substring beginning at index from and ending at index (to – 1). Note that s.substring(i,i+1) returns the character at index i.
String s = "example";
int i = 0;
// while there is an a in s
while (s.indexOf("a") >= 0)
{
// Find and save the next index for an a
i = s.indexOf("a");
// Process the string at that index
String ithLetter = s.substring(i,i+1);
...
}
The example in the mixed up code below finds and removes all the letter a’s in a string. You can watch it in action in this Java visualizer.
The following program removes all the a’s from a string, but the code is mixed up. Drag the blocks from the left area into the correct order in the right area. Click on the “Check Me” button to check your solution.
public static void main(String[] args)
{
---
String s = "are apples tasty without an a?";
int index = 0;
System.out.println("Original string: " + s);
---
// while there is an a in s
while (s.indexOf("a") >= 0)
{
---
// Find the next index for an a
index = s.indexOf("a");
---
// Remove the a at index by concatenating
// substring up to index and then rest of the string.
s = s.substring(0,index) +
s.substring(index+1);
---
} // end loop
---
System.out.println("String with a's removed:" + s);
---
} // end method
Google has been scanning old books and then using software to read the scanned text. But, the software can get things mixed up like using the number 1 for the letter l. Try the code below (and in the Java visualizer) to clean up scanning mistakes like this.
The following code loops through a string replacing all 1’s with l’s. Trace through the code below with a partner and explain how it works on the given message. You can run it line by line in the Java visualizer. Note that indexOf here can work repeatedly to find the next occurrence of a 1 because they are replaced as soon as they are found.
Change the code to add code for a counter variable to count the number of 1’s replaced in the message and print it out. Change the message to have more mistakes with 1’s to test it.
While loops are often used with strings when you are looking for a certain character or substring in a string and do not know how many times the loop needs to run. For loops are used when you know you want to visit every character.
String s = "example";
// loop through the string from 0 to length
for(int i=0; i < s.length(); i++)
{
String ithLetter = s.substring(i,i+1);
// Process the string at that index
...
}
The following main method has the correct code to count the number of e’s in a string, but the code is mixed up. Drag the blocks from the left area into the correct order in the right area. Click on the “Check Me” button to check your solution.
public static void main(String[] args)
{
---
String message = "e is the most frequent English letter.";
int count = 0;
---
for(int i=0; i < message.length(); i++)
{
---
if (message.substring(i,i+1).equalsIgnoreCase("e"))
---
count++;
---
}
---
System.out.println(count);
---
}
Here is a for-loop that creates a new string that reverses the string s. We start with a blank string sReversed and build up our reversed string in that variable by copying in characters from the string s. You can also run this code in this Java visualizer link or by clicking on the Code Lens button below.
What would happen if you started the loop at 1 instead? What would happen if you used <= instead of <? What would happen if you changed the order in which you added the ithLetter in line 12?