Leetcode143. Reorder List重排链表

给定一个单链表 L：L0→L1→…→Ln-1→Ln ，

将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…

你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例 1:

给定链表 1->2->3->4, 重新排列为 1->4->2->3.

示例 2:

给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.

思路：

根据题的意思，可以理解成，将链表分成两半，将后一半倒着插入第一半当中，

那么就好办了。

将后一半反转，再合并。

  class Solution {
  public:
	  void reorderList(ListNode* head)
	  {
		  if (head == NULL)
			  return;
		  int len = 0;
		  ListNode *head1 = head;
		  while (head1)
		  {
			  len++;
			  head1 = head1->next;
		  }
		  if (len <= 2)
			  return;
		  int half = len / 2;
		  int cnt = len - half;
		  head1 = head;
		  ListNode *last = NULL;
		  while (cnt)
		  {
			  cnt--;
			  last = head1;
			  head1 = head1->next;
		  }
		  last->next = NULL;
		  ListNode *head2 = NULL;
		  last = head1;
		  head1 = head1->next;
		  ///
		  last->next = NULL;
		  while (head1 !=NULL)
		  {
			  ListNode *node = head1 ->next;
			  head1->next = last;
			  last = head1;
			  head1 = node;
		  }
		  head2 = last;
		  while (head != NULL && head2 != NULL)
		  {
			  //cout << "2" << endl;
			  ListNode *next1 = head->next;
			  ListNode *next2 = head2 -> next;
			  head->next = head2;
			  ///
			  head2->next = next1;
			  head = next1;
			  head2 = next2;
		  }
	  }
  };