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 ...

Credits:
Special thanks to @aadarshjajodia for adding this problem and creating all test cases.

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比较直观的一个解法，将odd结点与even结点分到两个list中，然后再把这两个list连接起来。

odd结点的list的头就是输入list的头，even结点的list的头则是输入list头结点的下一个结点。

/**
 * 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)
            return NULL;
        ListNode *oddHead =  head;
        ListNode *oddCur = oddHead;
        ListNode *evenHead =head->next;
        ListNode *evenCur = evenHead;
        
        while(evenCur && evenCur->next)
        {
            oddCur->next = evenCur->next;
            oddCur = oddCur->next;
            
            evenCur->next = oddCur->next;
            evenCur = evenCur->next;
        }
        
        oddCur->next = evenHead;
        
        return oddHead;
        
        
    }
};