Find if the String is a number or floating point number - Regular Expression

Find if a string contains only digits?
?
1
2
3
4
5
public static boolean stringContainsOnlyDigits(String s) {
     Pattern p = Pattern.compile( "^[0-9]+$" );
     Matcher m = p.matcher( s );
     return m.find();
 }

Find if the String is a Floating Point number.
?
1
2
3
4
5
6
7
8
9
public static boolean stringIsFloatingPointNumber(String s) {
     Pattern p = Pattern.compile( "^[-+]?[0-9]*\\.?[0-9]+$" );
     Matcher m = p.matcher( s );
     boolean found = m.find();
     if(found) {
         System.out.println("found: " + m.group(0));
     }
     return found;
 }

The caveat here is the strings are long text and Double.parseDouble() will be out of range. Such problems can be solved only using Regular Expressions.

Comments

Post a Comment

Popular posts from this blog

Print staircase with both base and height equal to n

Consider a staircase of size n=4, as shown below:       #     ##   ### #### Observe that its base and height are both equal to n, and the image is drawn using # symbols and spaces. The last line is not preceded by any spaces. Write a program that prints a staircase of size n. public static void main(String[] args) { int n = 10; if(n < 1 || n > 100) { throw new IllegalArgumentException("N should be between 0 and 100"); } for (int i = 1; i <= n; i++) { System.out.println(Stream.concat(Stream.generate(() -> " ").limit(n-i), Stream.generate(() -> "#").limit(i)) .collect(Collectors.joining())); } } }
Read more

Find the maximum occurring character in given String ?

public static void main(String[] args) { System.out.println(maxOccurringChar("122333444455555666666")); } public static Character maxOccurringChar(String s) { Map<Character, Integer> charCountMap = new HashMap<>(); char[] array = s.toCharArray(); for (int i = 0; i < array.length; i++) { if(array[i] != ' ') { Integer occ = charCountMap.getOrDefault(array[i], 0); charCountMap.put(array[i], ++occ); } } Map.Entry<Character, Integer> maxEntry = charCountMap.entrySet().stream() .max(Map.Entry.comparingByValue()).get(); // Map.Entry<Character, Integer> maxEntry = getEntryWithMaxValue(charCountMap); System.out.println(String.format("Character %s occurs %s times.", maxEntry.ge...
Read more

Find the longest substring without repeating characters

Sliding Window is a subarray running on a large array, which is a collection of underlying elements. Here we use two pointers, one pointer A always selects the first char of the max substring, from each iteration of the string. The second pointer B runs ahead of pointer A, until a duplicate character is found or the end of the string is reached. public int lengthOfLongestSubstring(String s) { int a_pointer = 0; int b_pointer = 0; int max = 0; Set<Character> hashSet = new HashSet<>(); int i = 0; while(b_pointer < s.length()) { if(!hashSet.contains(s.charAt(b_pointer))) { hashSet.add(s.charAt(b_pointer)); b_pointer++; max = Math.max(hashSet.size(), max); } else { hashSet.remove(s.charAt(a_pointer)); a_pointer++; } } return max; }
Read more