Implement Even-odd Merge

Leetcode #328

EPI 8.10

Description:

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.

Idea:

设置oddHead和evenHead track各个不同的部分。

Code:

/**
 * 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) {
        ListNode dummyOdd(-1);
        ListNode dummyEven(-1);
        ListNode* oddHead=&dummyOdd, *evenHead=&dummyEven;
        int count=0;
        while(head!=NULL){
            ++count;
            if(count%2==1){
                oddHead->next=head;
                oddHead=oddHead->next;
            }
            else{
                evenHead->next=head;
                evenHead=evenHead->next;
            }
            head=head->next;
        }
        oddHead->next=dummyEven.next;
        evenHead->next=NULL;
        return dummyOdd.next;
    }
};