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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 ...
分析：题意为给一个单链表，将所有偶数位节点组合起来后面接上奇数位节点，要求算法的时间复杂度为O(n)空间复杂度为O(1)，注意：奇数位和偶数位组合内相对顺序不变，第一个节点被认为是偶节点，第二个节点为奇节点。。。
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) return head;
ListNode *odd,*even,*even_head;
odd=head;
even=head->next;
even_head=head->next;
while(even->next){
odd->next=even->next;
odd=odd->next;
if(odd->next){
even->next=odd->next;
even=even->next;
}
else{
even->next=NULL;
}
}
odd->next=even_head;
return head;
}
};
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原文地址：http://www.cnblogs.com/carsonzhu/p/5199270.html
