Strings and StringBuilder

Reading a string#

String s = "programming";
s.length()          // 11
s.charAt(0)         // 'p'
s.charAt(s.length() - 1)   // 'g'
s.isEmpty()         // false

Positions run from 0 to length() - 1. charAt(11) throws StringIndexOutOfBoundsException.

Pieces and searching#

s.substring(3, 7)       // "gram": from 3 up to, not including, 7
s.substring(7)          // "ming"
s.indexOf("gram")       // 3, or -1 if not found
s.indexOf('m')          // 6
s.contains("gram")      // true
s.startsWith("pro")     // true
s.endsWith("ing")       // true

Changing case, trimming, replacing#

"  Hi There  ".trim()          // "Hi There"
"Hi".toUpperCase()             // "HI"
"Hi".toLowerCase()             // "hi"
"banana".replace('a', 'o')     // "bonono"

Splitting and joining#

String[] parts = "a,b,c".split(",");             // ["a", "b", "c"]
String[] words = "one two   three".split("\\s+"); // any run of whitespace
String joined = String.join("-", "a", "b", "c");  // "a-b-c"

split takes a regular expression, which is why whitespace is written "\\s+" and a full stop must be escaped: split("\\.").

Strings never change#

Every method above returns a new string. The original is untouched:

String s = "hello";
s.toUpperCase();          // result thrown away
System.out.println(s);    // still "hello"
s = s.toUpperCase();      // keep the result

That is why building a string with + inside a loop is slow: each + creates a new string and copies everything so far. For a thousand pieces, that is a lot of copying.

StringBuilder#

When you build text piece by piece, use a StringBuilder, which can change:

StringBuilder sb = new StringBuilder();
for (int i = 0; i < 5; i++) {
    sb.append(i).append(' ');
}
String result = sb.toString();      // "0 1 2 3 4 "

new StringBuilder("hello").reverse().toString();   // "olleh"
sb.insert(0, "start ");
sb.deleteCharAt(sb.length() - 1);
sb.setCharAt(0, 'S');

reverse gives the quickest palindrome check: compare a string with its reverse using equals.

Comparing strings#

a.equals(b)                  // same characters: use this
a.equalsIgnoreCase(b)        // ignoring case
a.compareTo(b)               // negative, 0 or positive: alphabetical order

Never a == b for content: that asks whether they are the same object.

Walking the characters#

int vowels = 0;
for (char c : s.toCharArray()) {
    if ("aeiouAEIOU".indexOf(c) >= 0) vowels++;
}

Or by position, with s.charAt(i), when you need the index too.

Key points#

• length(), charAt(i), substring(a, b) and indexOf cover most reading.
• Strings never change; methods return new strings, so keep the result.
• Build text in loops with StringBuilder, then call toString().
• Compare with equals, never ==.

Next: decisions with if and switch.