New Problem "Odd Even Linked List"

haoel · haoel · commit 1ab20b068dac · 2016-01-16T22:03:01.000+08:00
diff --git a/README.md b/README.md
@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|328|[Odd Even Linked List](https://leetcode.com/problems/odd-even-linked-list/) | [C++](./algorithms/cpp/oddEvenLinkedList/OddEvenLinkedList.cpp)|Easy|
 |327|[Count of Range Sum](https://leetcode.com/problems/count-of-range-sum/) | [C++](./algorithms/cpp/countOfRangeSum/CountOfRangeSum.cpp)|Hard|
 |326|[Power of Three](https://leetcode.com/problems/power-of-three/) | [C++](./algorithms/cpp/powerOfThree/PowerOfThree.cpp)|Easy|
 |322|[Coin Change](https://leetcode.com/problems/coin-change/) | [C++](./algorithms/cpp/coinChange/coinChange.cpp)|Medium|
diff --git a/algorithms/cpp/oddEvenLinkedList/OddEvenLinkedList.cpp b/algorithms/cpp/oddEvenLinkedList/OddEvenLinkedList.cpp
@@ -0,0 +1,53 @@
+// Source : https://leetcode.com/problems/odd-even-linked-list/
+// Author : Hao Chen
+// Date   : 2016-01-16
+
+/*************************************************************************************** 
+ *
+ * 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 ...
+ * 
+ * Credits:Special thanks to @aadarshjajodia for adding this problem and creating all 
+ * test cases.
+ ***************************************************************************************/
+
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* oddEvenList(ListNode* head) {
+        if (!head) return head;
+        ListNode* pOdd = head;
+        ListNode* p = head->next;
+        ListNode* pNext = NULL;
+        while(p && (pNext=p->next)) {
+            
+            p->next = pNext->next;
+            pNext->next = pOdd->next;
+            pOdd->next = pNext;
+            
+            p = p->next;
+            pOdd = pOdd->next;
+        
+        }
+        return head;
+    }
+};
