Leetcode 141重排链表

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第一种思路：（时间：144ms    内存：12MB）

           1.指针找到中点

           2.拆成两个链表 ,拆成前后两条链line1,line2。逆置line2.

           3.遍历两个链表，后面的塞到前面的“缝隙里”。

节点为偶数时：size(line1)=size(line2)

节点为奇数时：size(line1)=size(line2)+1；

省去了每次找链尾的时间。当链表很长时效果明显。然而内存并没有明显改善。

第二种思路：双端队列

优点：简洁

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

//第一种思路
class Solution {
public:
	void reorderList(ListNode* head) {
		if (head == NULL || head->next == NULL) return;

		ListNode* p = head, *q;
		int count = 0;
		while (p != NULL)
		{
			p = p->next;
			count++;
		}
		p = head;
		//找中间位置
		for (int i = 1; i < (count +1) / 2; i++)//len(line1)>=len(line2)
			p = p->next;
		//拆成两条链,添加头结点
  		ListNode* line1 = new ListNode(-1);
		line1->next = head;
		ListNode* line2 = new ListNode(-1);
		line2->next = p->next;
		p->next = NULL;
		//逆置第二条链
		reverse(line2);

		p = line1->next;
		q = line2->next;
        
		while (p&&q)
		{
			ListNode* tmp = q;
			q = q->next;
			tmp->next = p->next;
			p->next = tmp;
			p = p->next->next;
		}
		head = line1->next;
		return;
	}

	void reverse(ListNode *head)
	{
		ListNode*p = head->next;
		head->next = NULL;
		while (p)
		{
			ListNode* q = p;//这里千万要记录p，否则p->next变成了NULL
			p = p->next;
			q->next = head->next;
			head->next =q;
		}
	}

};


//第二种思路
class Solution {
    public void reorderList(ListNode head) {
        LinkedList<ListNode> queue = new LinkedList<>();
        ListNode cur = head;
        while (cur != null) {
            queue.addLast(cur);
            cur = cur.next;
        }
        while (!queue.isEmpty()) {
            if (cur == null) {
                cur = queue.pollFirst();
            } else {
                cur.next = queue.pollFirst();
                cur = cur.next;
            }
            cur.next = queue.pollLast();
            cur = cur.next;
        }
        if (cur != null) {
            cur.next = null;
        }
    }
}