主题：求教大家一个算法问题 -- looklook

Yes, I put it too simply.

Here is how you can do it:

boolean hasDuplicate(int[] a) {

HashSet hs = new HashSet();

for (int i = 0; i < a.length; ++i) {

if (hs.contains(a[i]) {

return true;

}

else {

hs.add(a[i]);

}

}

return false;

}

Here is some Javadoc for HashSet class:

This class offers constant time performance for the basic operations (add, remove, contains and size), assuming the hash function disperses the elements properly among the buckets.

So, as you can see, the algorithm takes O(n) time. There is no need for sorting according to the problem requirement.

Of course, you can also define the method to return the actual duplicate number if you want.