【LeetCode 328】 Odd Even Linked List

题目描述：

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 1：

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL

Example 2：

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->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 …

思路：

遍历链表，每个节点的next指针指向该节点的next的next节点。

代码：

// 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 == NULL || head->next == NULL || head->next->next == NULL) return head;

        ListNode* steven = head;
        ListNode* odd = head->next;
        ListNode* now = head;

        int cnt = 1;
        while(head != NULL && head->next != NULL && head->next->next != NULL) {
            now = head->next;
            head->next = head->next->next;
            head = now;
            // cout << head->val << " " << endl;
            cnt++;
        }

        // cout << cnt << "===\n";
        cnt += 1;
        if (cnt % 2 !=  0) {
            ListNode *temp = head;
            head = head->next;
            // cout << temp->val << endl;
            temp->next = NULL;
        }
        head->next = odd;

        return steven;
    }
};

流下了没有技术的泪水。。。我这个逻辑啊，确实需要加强。。。