[leetcode] Sort List

最新推荐文章于 2022-02-24 15:47:33 发布

少林小钢炮

最新推荐文章于 2022-02-24 15:47:33 发布

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本文介绍了一种使用常数空间复杂度实现O(n log n)时间复杂度的链表排序算法。该算法通过快慢指针找到中间节点，将链表分为两部分并递归排序，最后合并两个有序链表。

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题目：

Sort List

Total Accepted: 9356 Total Submissions: 47695 My Submissions

Sort a linked list in O(n log n) time using constant space complexity.

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class Solution {
public:
	ListNode *sortList(ListNode *head)
	{
		if (head == NULL || head->next == NULL) return head;
		//第一步：从中间分开链表(使用快慢指针)
		ListNode * fast_p = head, *slow_p = head;
		while (fast_p->next != NULL && fast_p->next->next != NULL)
		{
			fast_p = fast_p->next->next;
			slow_p = slow_p->next;
		}
		//第二步：分别对两段链表排序（递归）
		ListNode * list1 = head;
		ListNode * list2 = slow_p->next; slow_p->next = NULL;
		list1 = sortList(list1);
		list2 = sortList(list2);
		//第三步：合并
		return merge(list1, list2);
	}
	ListNode * merge(ListNode * list1, ListNode * list2)
	{//合并两个已经排序的List
		if (list1 == NULL) return list2;
		if (list2 == NULL) return list1;
		ListNode dummy(-1);//小技巧
		ListNode * current = &dummy;
		while (list1 != NULL && list2 != NULL)
		{
			if (list1->val < list2->val)//每次从List1和List2中选择小的
			{
				current->next = list1; list1 = list1->next; current = current->next;
			}
			else
			{
				current->next = list2; list2 = list2->next; current = current->next;
			}
		}
		if (list1 != NULL)//如果List1还有节点
			current->next = list1;
		if (list2 != NULL)
			current->next = list2;

		return dummy.next;
	}
};