diff --git a/cpp/328_Odd_Even_Linked_List.cpp b/cpp/328_Odd_Even_Linked_List.cpp
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+// 328. Odd Even Linked List
+/**
+ * Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about 
+ * the node number and not the value in the nodes.
+ * 
+ * You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.
+ * 
+ * Example:
+ * Given 1->2->3->4->5->NULL,
+ * return 1->3->5->2->4->NULL.
+ * 
+ * Note:
+ * The relative order inside both the even and odd groups should remain as it was in the input. 
+ * The first node is considered odd, the second node even and so on ...
+ * 
+ * Tags: Linked List
+ * 
+ * Author: Kuang Qin
+ */
+
+#include <iostream>
+
+using namespace std;
+
+/**
+ * Definition for singly-linked list.
+ */
+struct ListNode {
+    int val;
+    ListNode *next;
+    ListNode(int x) : val(x), next(NULL) {}
+    ListNode(int x, ListNode *p) : val(x), next(p) {}
+};
+
+class Solution {
+public:
+    ListNode* oddEvenList(ListNode* head) {
+        if (head == NULL || head->next == NULL) {
+            return head;
+        }
+        
+        ListNode *odd = head, *even = head->next, *evenHead = even;
+        while (even != NULL && even->next != NULL) {
+            odd->next = odd->next->next;
+            even->next = even->next->next;
+            odd = odd->next;
+            even = even->next;
+        }
+        
+        odd->next = evenHead;
+        return head;
+    }
+};
+
+int main() {
+    ListNode node5(5), node4(4, &node5), node3(3, &node4), node2(2, &node3), node1(1, &node2);
+    Solution sol;
+    ListNode *newhead = sol.oddEvenList(&node1);
+    for (ListNode *p = newhead; p != NULL; p = p->next) {
+        cout << p->val << " ";
+    }
+    cout << endl;
+    cin.get();
+    
+    return 0;
+} 
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