LeetCode Sort List

最新推荐文章于 2022-02-24 15:47:33 发布

原创最新推荐文章于 2022-02-24 15:47:33 发布 · 237 阅读

0 ·

CC 4.0 BY-SA版权

面试题专栏收录该内容

144 篇文章

订阅专栏

本文介绍了一种使用分治法和合并排序对链表进行排序的方法，可在O(n log n)时间内完成排序，并保持常数级别的空间复杂度。通过快慢指针技术将链表分为两部分，递归地排序每部分后合并。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

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

代码思想参考http://blog.youkuaiyun.com/xudli/article/details/16819545

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *sortList(ListNode *head) {
        return mergeSort(head);
    }
private:
	ListNode *mergeSort(ListNode *head) {
		if(head == NULL || head->next == NULL)
			return head; 
		ListNode *slow = head, *fast = head;
		ListNode *pre_slow = head;
		//分治法：使用快慢指针进行分组，slow之前为一组，之后为一组；
		while(fast != NULL && fast->next != NULL) {
			fast = fast->next->next;
			pre_slow = slow;
			slow = slow->next;
		}
		pre_slow->next = NULL;
		ListNode *h1 = mergeSort(head);
		ListNode *h2 = mergeSort(slow);
		return merge(h1, h2);	
	}
	ListNode *merge(ListNode *h1, ListNode *h2) {
		ListNode *dummy = new ListNode(-1);
		ListNode *cur = dummy;
		if(h1 == NULL && h2 == NULL)
			return NULL;
		while(h1 != NULL && h2 != NULL) {
			if(h1->val <= h2->val) {
				cur->next = h1;
				h1 = h1->next;
			}
			else {
				cur->next = h2;
				h2 = h2->next;
			}
			cur = cur->next;
		}
		//链表h1还有剩余 
		while(h1 != NULL) {
			cur->next = h1;
			h1 = h1->next;
			cur = cur->next;
		}
		//链表h2有剩余 
		while(h2 != NULL) {
			cur->next = h2;
			h2 = h2->next;
			cur = cur->next;
		} 
		cur = dummy->next;
		delete dummy;
		return cur; 	
	}
};