328. Odd Even Linked List --M

Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.

The first node is considered odd, and the second node is even, and so on.

Note that the relative order inside both the even and odd groups should remain as it was in the input.

You must solve the problem in O(1) extra space complexity and O(n) time complexity.

 

Example 1:

Input: head = [1,2,3,4,5]
Output: [1,3,5,2,4]

Example 2:

Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]

 

Constraints:

• The number of nodes in the linked list is in the range [0, 104].
• -106 <= Node.val <= 106

A:

-------------------------就是一个个拆下来而已------------

/**

* Definition for singly-linked list.

* struct ListNode {

* int val;

* ListNode *next;

* ListNode() : val(0), next(nullptr) {}

* ListNode(int x) : val(x), next(nullptr) {}

* ListNode(int x, ListNode *next) : val(x), next(next) {}

* };

*/

class Solution {

public:

ListNode* oddEvenList(ListNode* head) {

ListNode oHeader, eHeader;

auto oTail = &oHeader, eTail = &eHeader;

while (head) {

oTail->next = head;

oTail = oTail->next;

head = head->next;

if (head) {

eTail->next = head;

head = head->next;

eTail = eTail->next;

eTail->next = nullptr;

}

}

oTail->next = eHeader.next;

return oHeader.next;

}

};

Mistakes:           
ListNode* header1(0);         这样是不行的，其实并没有真正的声明一个pointer

正确的方法：
 ListNode header1();
或者    ListNode* header1 = new ListNode(0); ----> declared on heap, need delete