58-Odd Even Linked List

1. Odd Even Linked List My Submissions QuestionEditorial Solution
   
   Total Accepted: 29496 Total Submissions: 77710 Difficulty: Medium
   
   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 …

思路：主要是操作指针指向

时间：O(n) 节点数n

空间：O(1)

/**
 * 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;//头部为空或1个
        ListNode * oddhead = head,*evenhead=head->next;
        ListNode *op=oddhead,*ep=evenhead,*opend=oddhead;//奇偶链首节点
        while(op!=NULL&&ep!=NULL){  //循环修改指向
            op->next = ep->next;
            if(op->next!=NULL)ep->next = op->next->next;
            opend = op;
            if(op!=NULL)op = op->next;
            if(ep!=NULL)ep = ep->next;
        }
        if(op!=NULL)opend=op;
        opend->next = evenhead;
        return head;
    }
};