[en] Add Partition Array by Odd and Even

Author: JuneYuan

en/integer_array/partition_array_by_odd_and_even.md

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+# Partition Array by Odd and Even
+
+## Question
+
+- lintcode: [(373) Partition Array by Odd and Even](http://www.lintcode.com/en/problem/partition-array-by-odd-and-even/)
+- [Segregate Even and Odd numbers - GeeksforGeeks](http://www.geeksforgeeks.org/segregate-even-and-odd-numbers/)
+
+```
+Partition an integers array into odd number first and even number second.
+
+Example
+Given [1, 2, 3, 4], return [1, 3, 2, 4]
+
+Challenge
+Do it in-place.
+```
+
+## Solution
+
+Use **two pointers** to keep the odd before the even, and swap when necessary.
+
+### Java
+
+```java
+public class Solution {
+    /**
+     * @param nums: an array of integers
+     * @return: nothing
+     */
+    public void partitionArray(int[] nums) {
+        if (nums == null) return;
+
+        int left = 0, right = nums.length - 1;
+        while (left < right) {
+            // odd number
+            while (left < right && nums[left] % 2 != 0) {
+                left++;
+            }
+            // even number
+            while (left < right && nums[right] % 2 == 0) {
+                right--;
+            }
+            // swap
+            if (left < right) {
+                int temp = nums[left];
+                nums[left] = nums[right];
+                nums[right] = temp;
+            }
+        }
+    }
+}
+```
+
+### C++
+
+```c++
+void partitionArray(vector<int> &nums) {
+      if (nums.empty()) return;
+
+      int i=0, j=nums.size()-1;
+      while (i<j) {
+          while (i<j && nums[i]%2!=0) i++;
+          while (i<j && nums[j]%2==0) j--;
+          if (i != j) swap(nums[i], nums[j]);
+      }
+  }
+```
+
+### Src Code Analysis
+
+Be careful not to forget `left < right` in while loop condition.
+
+### Complexity
+
+To traverse the array, time complexity is $$O(n)$$. And maintain two pointers mean $$O(1)$$ space complexity.