【算法刷题】leetcode reorder-list

Neo_dot

于 2018-07-04 19:26:43 发布

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本文介绍了一种在不改变节点值的情况下，对单链表进行重新排序的方法。具体实现包括找到链表中点、反转后半部分链表，然后交替合并前半部分链表与反转后的后半部分链表。

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Given a singly linked list L: L 0→L 1→…→L n-1→L n,
reorder it to: L 0→L n →L 1→L n-1→L 2→L n-2→…

You must do this in-place without altering the nodes' values.

For example,
Given{1,2,3,4}, reorder it to{1,4,2,3}.

	void reorderList(ListNode *head) {
		
		if (head == nullptr)
			return;
		//fast and low point to find the middle pos 
		ListNode* slow = head;
		ListNode* fast = head;

		if (fast->next == nullptr)
			return;

		fast = fast->next;

		if (fast->next == nullptr)
			return;
		
		fast = fast->next;

		while (fast->next != nullptr)
		{
			fast = fast->next;
			
			if (fast->next != nullptr)
				fast = fast->next;

			slow = slow->next;
		}

		//split and reverse

		ListNode* after = slow->next;
		slow->next = nullptr;
		ListNode* pre = nullptr;
		while (after !=nullptr)
		{
			ListNode* temp = after->next;
			if (temp == nullptr)
				fast = after;

			after->next = pre;
			pre = after;
			after = temp;

		} 
		//merge fast and slow, slow is first

		slow = head;
		while (slow != nullptr && fast != nullptr)
		{
			pre = fast;
			fast = fast->next;
			pre->next = slow->next;
			slow->next = pre;
			slow = pre->next;			
		}
        //only fast list can more than one
        pre->next = fast;



	}