I want to find the first repeated character from a string. I usually do it using array_intersect in php. Is there something similar in Java?
For example:
String a=zxcvbnmz
Desired output : z
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What do you actually need to get those values for? Maybe we can answer that question.Steve Smith– Steve Smith2017-05-12 12:39:45 +00:00Commented May 12, 2017 at 12:39
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Been trying to learn java and there's this question to display the first repeated character in a string. It's rather simple in php. Is there an inbuilt function I could use to simplify the solution?Red Bottle– Red Bottle2017-05-12 12:41:48 +00:00Commented May 12, 2017 at 12:41
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2 Answers
array_intersect — Computes the intersection of arrays (source)
So in this case you can use Set::retainAll :
Integer[] a = {1,2,3,4,5};
Integer[] b = {2,4,5,6,7,8,9};
Set<Integer> s1 = new HashSet<>(Arrays.asList(a));
Set<Integer> s2 = new HashSet<>(Arrays.asList(b));
s1.retainAll(s2);
Integer[] result = s1.toArray(new Integer[s1.size()]);
System.out.println(Arrays.toString(result));
Output
[2, 4, 5]
You can read about this here Java, find intersection of two arrays
1 Comment
Red Bottle
Ive upvoted sir. This is totally legit. Idk why its downvoted
There's no default implementation for this behavior; however, you can code your own solution! Since you want to find the first repeated character, you can make a HashSet of Characters. As you iterate through the array, you add each character to the HashSet until you come across a character already in the HashSet - this must be the first repeated character. Example code below:
public char arrayIntersect(String string) {
HashSet<Character> hashSet = new HashSet<>();
for (int i = 0; i < string.length(); i++) {
char c = string.charAt(i);
if (hashSet.contains(c))
return c;
else
hashSet.add(c);
}
return null;
}
This runs in O(n) time, as HashSet lookups run in O(1) time.
