Reverse Operations !!!!!!!!!!!! (Facebook 练习题)

You are given a singly-linked list that contains N integers. A subpart of the list is a contiguous set of even elements, bordered either by the end of the list or an odd element. For example, if the list is [1, 2, 8, 9, 12, 16], the subparts of the list are [2, 8] and [12, 16].

Then, for each subpart, the order of the elements is reversed. In the example, this would result in the new list, [1, 8, 2, 9, 16, 12].

The goal of this question is: given a resulting list, determine the original order of the elements.

Implementation detail:

You must use the following definition for elements in the linked list:

class Node {
    int data;
    Node next;
}

Signature Node reverse(Node head)

Constraints 1 <= N <= 1000, where N is the size of the list 1 <= Li <= 10^9, where Li is the ith element of the list

Example Input: N = 6 list = [1, 2, 8, 9, 12, 16]

Output: 

[1, 8, 2, 9, 16, 12] 

A:

哎，这个题，错了好多个地方， 真是不应该啊不应该

// struct Node {
//   int data;
//   Node *next;
//   Node(int x) : data(x), next(NULL) {}
// };


Node* reverse(Node *head) {
  // Write your code here
  Node * header =  new Node(0);
  Node * tail = header;
  
  while(head){    
    Node * tmpHeader =  new Node(0);
    Node * tmpTail = head;    
    
    if(head->data %2 == 1){ // if even number
        Node * tmp = head; 
        head = head->next;
        tmp->next = tmpHeader->next;
        tmpHeader->next = tmp;
    }else{
      while(head && head->data %2 == 0){   
        Node * tmp = head; 
        head = head->next;
        tmp->next = tmpHeader->next;
        tmpHeader->next = tmp;
      }
    }
    tail->next = tmpHeader->next;
    tail = tmpTail;    
  }
  
  return header->next;
}

Mistakes：

1:  *************

Node * header =  new Node(0);  

而写成Node * header（0） // 是不对的。 segment fault