Leetcode-328. Odd Even Linked List

题目

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 …
注意奇偶是节点位置为奇偶而不是节点的值

思路

创建两个结果链表，一次遍历，将奇数节点追加到第一个链表之后，偶数节点追加到第二个节点之后，最后拼接两个链表即可。

代码

class Solution {public:    ListNode* oddEvenList(ListNode* head) {        if(NULL == head || head->next == NULL)            return head;        ListNode *h1 = head, *h2 = head->next;        ListNode *p1 = h1, *p2 = h2, *p = head->next->next;        bool flag = true;        while(p) {            if(flag) {                p1->next = p;                p1 = p1->next;            } else {                p2->next = p;                p2 = p2->next;            }            p = p->next;            flag = !flag;        }        p2->next = NULL;        p1->next = h2;        return h1;    }};