upload 1 answer, Aug 17th

Author: stevenlordiam

Medium/SingleNumberIII.java

@@ -0,0 +1,41 @@
+/*
+Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.
+
+For example:
+Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].
+
+Note:
+The order of the result is not important. So in the above example, [5, 3] is also correct.
+Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?
+*/
+
+public class SingleNumberIII {
+	/*
+	先将所有的数异或，得到的将是x和y以后之后的值n。 因为这两个数一定是不同的，所以最终异或的值至少有一个位为1
+    找到这个数n的为1的某一位（为了方便就取最右边为1的一位， n & ~(n-1))，然后根据该位的值是否为1，将数组中的每一个数，分成两个部分
+    这样每个部分，就可以采用Sing number I中的方法得到只出现一次的数
+	*/
+    public int[] singleNumber(int[] nums) {
+        int[] res = new int[2];
+        int n = 0;
+        for (int elem : nums) {
+            n ^= elem;
+        }
+        n = n & (~(n-1));
+        for (int elem : nums) {
+            if ((elem & n) != 0) {
+                res[0] = res[0]^elem;
+            }
+            else res[1] = res[1]^elem;
+        }
+        return res;
+    }
+}
+
+/*
+https://leetcode.com/discuss/52351/accepted-java-space-easy-solution-with-detail-explanations
+http://www.cnblogs.com/EdwardLiu/p/4391455.html
+https://richdalgo.wordpress.com/2015/02/01/lintcode-single-number-iii/
+https://github.com/shawnfan/LintCode/blob/master/Java/Single%20Number%20III.java
+http://www.wengweitao.com/lintcode-single-number-i-ii-iii-luo-dan-de-shu.html
+*/