150。3 - sort List- AC

最新推荐文章于 2024-08-10 09:07:16 发布

kingoverlord

最新推荐文章于 2024-08-10 09:07:16 发布

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本文详细解析了一种基于链表的排序算法实现过程，通过寻找中间节点将链表分为两部分，递归地对每部分进行排序，最后将两部分有序链表合并为一个有序链表。文章提供了完整的代码实现，包括查找中间节点、合并两个有序链表等关键步骤。

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终于 ac 一道题···

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
ListNode * findMid(ListNode *phead)
{
	if (phead->next->next == NULL || phead->next == NULL || phead == NULL)
	{
		return phead;
	}
	ListNode * pslow = phead;
	ListNode * pquick = phead;

	while (pquick->next && pquick->next->next)// the last second is not visited; after this process the "pquick" point to the last second;
	{
		pslow = pslow->next;
		pquick = pquick->next->next;
	}
	return pslow;
}
ListNode *Merge(ListNode *ph1, ListNode *ph2)
{
	if (ph1 == NULL) return ph2;
	if (ph2 == NULL) return ph1;
	ListNode *p1 = ph1;
	ListNode *p2 = ph2;
	ListNode *pnew = 0;
	if (p1->val > p2->val)
	{
		pnew = p2;
		p2 = p2->next;
	}
	else
	{
		pnew = p1;
		p1 = p1->next;
	}
	ListNode *pres = pnew;

	while (p1 && p2)
	{
		if (p1->val < p2->val)
		{
			pnew->next = p1;
			p1 = p1->next;
		}
		else
		{
			pnew->next = p2;
			p2 = p2->next;
		}
		pnew = pnew->next;
	}
	p1 == NULL ? (pnew->next = p2) : (pnew->next = p1);
	return pres;
}
ListNode * sortList(ListNode *head)
{
	if (head == NULL || head->next == NULL) return head;
	ListNode *mid = findMid(head);
	ListNode *head2 = mid->next;
	mid->next = NULL;
 ListNode *ph1 = sortList(head);
	ListNode *ph2 = sortList(head2);
	return Merge(ph1, ph2);
	//Merge(sortList(head), sortList(head2));
}
};