New Problem Solution - "Sort Array By Parity"

README.md

@@ -29,6 +29,7 @@ LeetCode
 |941|[Valid Mountain Array](https://leetcode.com/problems/valid-mountain-array/) | [Python](./algorithms/python/ValidMountainArray/validMountainArray.py)|Easy|
 |931|[Minimum Falling Path Sum](https://leetcode.com/problems/minimum-falling-path-sum/) | [C++](./algorithms/cpp/minimumFallingPathSum/MinimumFallingPathSum.cpp)|Medium|
 |914|[X of a Kind in a Deck of Cards](https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/) | [Python](./algorithms/python/XOfAKindInADeckOfCards/hasGroupsSizeX.py)|Easy|
+|905|[Sort Array By Parity](https://leetcode.com/problems/sort-array-by-parity/) | [C++](./algorithms/cpp/sortArrayByParity/SortArrayByParity.cpp)|Easy|
 |859|[Buddy Strings](https://leetcode.com/problems/buddy-strings/description/) | [C++](./algorithms/cpp/buddyStrings/BuddyStrings.cpp)|Easy|
 |858|[Mirror Reflection](https://leetcode.com/problems/mirror-reflection/description/) | [C++](./algorithms/cpp/mirrorReflection/MirrorReflection.cpp)|Medium|
 |852|[Peak Index in a Mountain Array](https://leetcode.com/problems/peak-index-in-a-mountain-array/description/) | [C++](./algorithms/cpp/peakIndexInAMountainArray/PeakIndexInAMountainArray.cpp)|Easy|

algorithms/cpp/sortArrayByParity/SortArrayByParity.cpp

@@ -0,0 +1,41 @@
+// Source : https://leetcode.com/problems/sort-array-by-parity/
+// Author : Hao Chen
+// Date   : 2019-03-26
+
+/***************************************************************************************************** 
+ *
+ * Given an array A of non-negative integers, return an array consisting of all the even elements of 
+ * A, followed by all the odd elements of A.
+ * 
+ * You may return any answer array that satisfies this condition.
+ * 
+ * Example 1:
+ * 
+ * Input: [3,1,2,4]
+ * Output: [2,4,3,1]
+ * The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.
+ * 
+ * Note:
+ * 
+ * 	1 <= A.length <= 5000
+ * 	0 <= A[i] <= 5000
+ * 
+ ******************************************************************************************************/
+
+class Solution {
+public:
+    bool isEven(int& x) {
+        return x % 2 == 0;
+    }
+    vector<int> sortArrayByParity(vector<int>& A) {
+        //two pointer, one from left to right, another from right to left
+        // if left is odd number and right is even number, switch them
+        int l=0, r=A.size()-1;
+        while ( l < r ) {
+            if ( !isEven(A[l]) && isEven(A[r]) ) swap(A[l], A[r]);
+            if ( isEven(A[l]) ) l++;
+            if ( !isEven(A[r]) ) r--;
+        }
+        return A;
+    }
+};