*single number

Given An array of numbers nums , in which exactly-elements appear only once and all the other elements appear exactly Twice. Find the elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5] , return [3, 5] .

Note:

1. The order of the result is not important. The above example, is [5, 3] also correct.
2. Your algorithm should run in linear runtime complexity. Could implement it using only constant space complexity?

Credits:
Special thanks to @jianchao. Li.fighter for adding the problem and creating all test cases.

1 /**2 * This code is provided by the nine-chapter algorithm editor. No copyright welcome forwarding. 3 *-Nine chapters The algorithm is committed to helping more Chinese to find a good job, the teachers team from Silicon Valley and the domestic first-line large companies in-service engineers. 4 *-existing interview training courses include: Nine Chapters algorithm class, System design class, bat domestic class5 *-For more details, please visit the official website:http://www.jiuzhang.com/6  */7 8  Public classSolution {9     /**Ten      * @paramA:an Integer Array One      * @return: Integers A      */ -      PublicList<integer> SINGLENUMBERIII (int[] A) { -         intXOR = 0; the          for(inti = 0; i < a.length; i++) { -XOR ^=A[i]; -         } -          +         intLastbit = XOR-(XOR & (xor-1)); -         intGroup0 = 0, group1 = 0; +          for(inti = 0; i < a.length; i++) { A             if((Lastbit & a[i]) = = 0) { atGroup0 ^=A[i]; -}Else { -Group1 ^=A[i]; -             } -         } -          inarraylist<integer> result =NewArraylist<integer>(); - Result.add (group0); to Result.add (group1); +         returnresult; -     } the}

*single number