theveteranofjava
diff --git a/‎Java/2 Sum.java‎
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diff --git a/‎Java/3 Sum Closest.java‎
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diff --git a/‎Java/3 Sum Smaller.java‎
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diff --git a/‎Java/3 Sum.java‎
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diff --git a/‎Java/A+B.java‎
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diff --git a/‎Java/Add Binary.java‎
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diff --git a/‎Java/Add Two Numbers II.java‎
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diff --git a/‎Java/Add Two Numbers.java‎
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diff --git a/‎Java/Array Partition I.java‎
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diff --git a/‎Java/Backpack II.java‎
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@@ -41,25 +41,26 @@ Always check (target - current) from the HashMap.
     (key, value) = (numbers[i], i)
     Note: return index+1 because this is not 0-based.
 */
-
-public class Solution {
-     //Using HashMap
-    public int[] twoSum(int[] numbers, int target) {
-        if (numbers == null || numbers.length == 0) {
+class Solution {
+    final static int RESULT_SIZE = 2;
+    final static int RESULT_INDEX1 = 0;
+    final static int RESULT_INDEX2 = 1;
+    public int[] twoSum(int[] nums, int target) {
+        if (nums == null || nums.length == 0) {
             return null;
         }
-        int[] rst = new int[2];
-        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
-        for (int i = 0; i < numbers.length; i++) {
-            if (map.containsKey(target - numbers[i])) {
-                rst[0] = map.get(target - numbers[i]) + 1;
-                rst[1] = i + 1;
-                break;
+        final int[] result = new int[RESULT_SIZE];
+        final Map<Integer, Integer> recordMap = new HashMap<>();
+        for (int i = 0; i < nums.length; i++) {
+            if (recordMap.containsKey(target - nums[i])) {
+                result[RESULT_INDEX1] = recordMap.get(target - nums[i]);
+                result[RESULT_INDEX2] = i;
+                return result;
             } else {
-                map.put(numbers[i], i);
+                recordMap.put(nums[i], i);
             }
         }
-        return rst;
+        return result;
     }
 }
 
 
@@ -0,0 +1,32 @@
+E
+
+从结果出发, 只需要找到加法的结果，而不强调具体配对。找到排列取单数位的规律，再考虑负数和正数的相同规律，即可找到排列求解的方法。
+
+
+```
+/*
+Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.
+
+Example 1:
+Input: [1,4,3,2]
+
+Output: 4
+Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).
+Note:
+n is a positive integer, which is in the range of [1, 10000].
+All the integers in the array will be in the range of [-10000, 10000].
+*/
+
+class Solution {
+    public int arrayPairSum(int[] nums) {
+        Arrays.sort(nums);
+        int result = 0;
+        for (int i = 0; i < nums.length; i++) {
+            if (i % 2 == 0) {
+                result += nums[i];
+            }
+        }
+        return result;
+    }
+}
+```