add 905

README.md

@@ -291,3 +291,4 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 661 |[Image Smoother](https://leetcode.com/problems/image-smoother)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/661.%20Image%20Smoother.md)|Easy|  |
 | 665 |[Non-decreasing Array](https://leetcode.com/problems/non-decreasing-array)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/665.%20Non-decreasing%20Array.md)|Easy|  |
 | 714 |[Best Time to Buy and Sell Stock with Transaction Fee](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/714.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Transaction%20Fee.md)|Medium|  |
+| 905 |[Sort Array By Parity](https://leetcode.com/problems/sort-array-by-parity/)|[C++](solutions/905.%20Sort%20Array%20By%20Parity.md)|Medium|  |

solutions/905. Sort Array By Parity.md

@@ -0,0 +1,43 @@
+# [905. Sort Array By Parity](https://leetcode.com/problems/sort-array-by-parity/)
+
+# 思路
+
+题目要求将一个数组中的偶数放在奇数前面。很简单用两个指针i和j就可以了，分别表示`a[0, 1, ..., i-1]`全部为偶数而`a[j+1, j+2, ..., n-1]`全部为奇数，所以i和j初始分别为0和n-1。然后就是循环，每个循环内部：
+* 1. 如果`A[i]`为奇数那么交换`A[i]`和`A[j]`然后将`j--`，否则`i++`；
+* 2. 如果`A[j]`为偶数那么交换`A[i]`和`A[j]`然后将`i++`，否则`j--`；
+
+上面的1和2单独使用也是可以的。
+
+另外STL中`partition`所干的就是此题要求的事，它会使所有使给定比较函数返回 true 的元素放在返回false的前面，所以我们只需要自定义比较函数就可以了。
+
+# C++
+``` C++
+class Solution {
+public:
+    vector<int> sortArrayByParity(vector<int>& A) {
+        int i = 0, j = A.size() - 1;
+        
+        // 如果问题条件变一下, 例如将能被3整除的放在不能整除的前面, 我们只需要改这一行代码
+        # define CONDITION(i) (A[i] & 1)
+        
+        while(i < j){
+            if(i < j && CONDITION(i)) swap(A[i], A[j--]);
+            else i++;
+            
+            if (i < j && !CONDITION(j)) swap(A[i++], A[j]);
+            else j--;
+        }
+        return A;
+    }
+};
+```
+## 直接使用STL中的partition
+``` C++
+class Solution {
+public:
+    vector<int> sortArrayByParity(vector<int>& A) {
+        partition(A.begin(), A.end(), [](int a) { return !(a & 1);});
+        return A;
+    }
+};
+```