add problem 328 add test function

Author: supercoderhawk

C/301-350/328-Odd-Even-Linked-List.c

@@ -0,0 +1,61 @@
+/*
+ * 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 ...
+ */
+#include 
+#include "../headers/problem_301_350.h"
+
+struct ListNode *oddEvenList(struct ListNode *head) {
+    if (head == NULL || head->next == NULL || head->next->next == NULL) {
+        return head;
+    }
+
+    struct ListNode *pOddHead = head, *pEvenHead = head->next;
+    struct ListNode *pOdd = pOddHead, *pEven = pEvenHead, *p = pEven->next;
+
+    int count = 3;
+    while (p!=NULL) {
+        if (count % 2) {
+            pOdd->next = p;
+            pOdd = pOdd->next;
+        } else {
+            pEven->next = p;
+            pEven = pEven->next;
+        }
+        p = p->next;
+        count++;
+    }
+
+    pEven->next = NULL;
+    pOdd->next = pEvenHead;
+    return pOddHead;
+}
+
+void testOddEvenList() {
+    int a1[4] = {1, 2, 3, 4};
+    int a2[2] = {1, 2};
+    int a3[4] = {1};
+    struct ListNode *p;
+
+    p = createList(a1, 4);
+    printList(oddEvenList(p));
+
+    p = createList(a2, 2);
+    printList(oddEvenList(p));
+
+    p = createList(a3, 1);
+    printList(oddEvenList(p));
+
+    p = NULL;
+    printList(oddEvenList(p));
+
+}