public class Solution {public int[] singleNumber(int[] nums) {int[] res = {-1, -1};if(nums == null || nums.length == 0) {return res;}int a_XOR_b = 0;  for(int i = 0; i < nums.length; i++) {      // find a ^ ba_XOR_b ^= nums[i];}int k = 0;                              // find one bit in a^b == 1for(int i = 0; i < 31; i++) {           // that means only one num in a, b has 1 on kth bitif(((a_XOR_b >> i) & 1) == 1) {k = i;break;}}int a = 0;for(int i = 0; i < nums.length; i++) {   // since only one num in a, b has 1 on kth bitif(((nums[i] >> k) & 1) == 1) {      // we can XOR all nums has 1 on kth bita ^= nums[i];                    // duplicate nums will be evened out
            }}int b = a ^ a_XOR_b;res[0] = a;res[1] = b;return res;}
}

public class Solution {public int[] singleNumber(int[] nums) {int[] res = {-1, -1};if (nums == null || nums.length == 0) {return res;}int a_XOR_b = 0;for (int i : nums) {a_XOR_b ^= i;}int k = 0;for (int i = 0; i < 32; i++) {if (((a_XOR_b >> i) & 1) == 1) {k = i;break;}}res[0] = 0;for (int i : nums) {if (((i >> k) & 1) == 1) {res[0] ^= i;}}res[1] = a_XOR_b ^ res[0];return res;}
}

public class Solution {public int[] singleNumber(int[] nums) {int[] res = new int[2];if (nums == null || nums.length < 2) return res;int a_xor_b = 0;for (int num : nums) { a_xor_b ^= num; }int diffIndex = 0;while (diffIndex < 32) {if (((a_xor_b >> diffIndex) & 1) == 1) { break; }diffIndex++;}int num1 = 0;for (int num : nums) {if (((num >> diffIndex) & 1) == 1) { num1 ^= num; }}res[0] = num1;res[1] = a_xor_b ^ num1;return res;}
}

public class Solution {public int[] singleNumber(int[] nums) {if (nums == null || nums.length < 2) return new int[] {-1, -1};int a_XOR_b = 0;for (int num : nums) a_XOR_b ^= num;int k = 0;while (k < 31) {if (((a_XOR_b >> k) & 1) == 1) break;k++;}int a = 0;for (int num : nums) {if (((num >> k) & 1) == 1) a ^= num;}return new int[] {a, a ^ a_XOR_b};}
}