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;}}};