As mentioned by others, your algorithm is O(n^2). Here is an O(N) algorithm, because HashSet#add runs in constant time ( the hash function disperses the elements properly among the buckets) - Note that I originally size the hashset to the maximum size to avoid resizing/rehashing:
public static int findDuplicate(String s) {
char[] chars = s.toCharArray();
Set<Character> uniqueChars = new HashSet<Character> (chars.length, 1);
for (int i = 0; i < chars.length; i++) {
if (!uniqueChars.add(chars[i])) return i;
}
return -1;
}
Note: this returns the index of the first duplicate (i.e. the index of the first character that is a duplicate of a previous character). To return the index of the first appearance of that character, you would need to store the indices in a Map<Character, Integer>
(Map#put
is also O(1) in this case):
public static int findDuplicate(String s) {
char[] chars = s.toCharArray();
Map<Character, Integer> uniqueChars = new HashMap<Character, Integer> (chars.length, 1);
for (int i = 0; i < chars.length; i++) {
Integer previousIndex = uniqueChars.put(chars[i], i);
if (previousIndex != null) {
return previousIndex;
}
}
return -1;
}
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