add solution of problem 260: single number iii

Author: ppulse

SingleNumberIII260/README.md

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

SingleNumberIII260/Solution.java

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+public class Solution {
+    public int[] singleNumber(int[] nums) {
+        int[] result = new int[2];
+        if (nums == null || nums.length == 0)
+            return result;
+        
+        int xorResult = nums[0];
+        for (int i = 1; i < nums.length; i++)
+            xorResult ^= nums[i];
+        
+        int partitionNum = xorResult & (xorResult ^ (xorResult - 1));
+        List<List<Integer>> twoParts = partition(nums, partitionNum);
+        result[0] = singleNumberOneValue(twoParts.get(0));
+        result[1] = singleNumberOneValue(twoParts.get(1));
+        
+        return result;
+    }
+    
+    private List<List<Integer>> partition(int[] nums, int partitionNum) {
+        List<List<Integer>> result = new ArrayList<List<Integer>>();
+        List<Integer> partOne = new ArrayList<Integer>();
+        List<Integer> partTwo = new ArrayList<Integer>();
+        
+        for (int i = 0; i < nums.length; i++) {
+            if ((nums[i] & partitionNum) == 0)
+                partOne.add(nums[i]);
+            else
+                partTwo.add(nums[i]);
+        }
+        
+        result.add(partOne);
+        result.add(partTwo);
+        
+        return result;
+    }
+    
+    private int singleNumberOneValue(List<Integer> nums) {
+        int result = 0;
+        Iterator<Integer> iter = nums.iterator();
+        while (iter.hasNext())
+            result ^= iter.next();
+        return result;
+    }
+}