Odd Even Linked List问题及解法

问题描述：

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.

示例:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.

问题分析：

利用双指针，分别记录奇偶索引的最后一个值得指针，然后进行删除和插入操作即可。

过程详见代码：

/** * 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 == nullptr) return nullptr;ListNode* odd = head,*even = head->next,* t,* h;        if(!even) return head;        h = even->next;while (h){even->next = h->next;h->next = odd->next;odd->next = h;even = even->next;                odd = odd->next;if (even) h = even->next;else break;}return  head;    }};