Leetcode 328:Odd Even Linked List

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 ...

//给定一个单链表，将其节点进行分组，使得所有的奇数节点排列在前，偶数节点在后。
//请注意这里的奇偶指的是节点序号而不是节点的值。
//第一个节点为奇数节点，第二个节点为偶数节点，以此类推。
//你应当尝试“就地”完成此问题。程序应当满足O(1)的空间复杂度和O(nodes)的时间复杂度。
class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if(!head)  return head;
        ListNode* odd=head;
        ListNode* evenhead=head->next;//记录下第一个偶结点的位置
        ListNode* even=evenhead;
        while(even && even->next)
        {
            odd->next=odd->next->next;
            even->next=even->next->next;
            odd=odd->next;
            even=even->next;
        }
        odd->next=evenhead; //不要忘记将最后一个奇数节点与第一偶数节点链接
        return head;
    }
};