Problem:
given an array that contains duplicates (except one value), find the one value that does not have a duplicate in that array. Explain the complexity of your algorithm. So in the array: A = [2, 3, 4, 5, 2, 3, 4] your algorithm should return 5 since that‘s the value that does not have a duplicate.
Solution:
this can be solved using a HashMap (two pass through). Or 2 HashSets (1 pass through).
I will show the 2nd solution (2 hashset)
import java.util.*; public class Dup{ public static void main(String[] args){ int[] n= {2,3,4,5,2,3,4,1,1,1}; Set<Integer> all= new HashSet<>(); Set<Integer> unq= new HashSet<>(); for(int i:n){ if(all.contains(i)){ unq.remove(i); } else{ all.add(i); unq.add(i); } } Iterator i= unq.iterator(); while(i.hasNext()){ System.out.println(""+ i.next()); } } }
find unique values in an array
原文：http://www.cnblogs.com/miaoz/p/4098647.html