LeetCode 148. Sort List (排序列表)

最新推荐文章于 2025-05-22 22:16:20 发布

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最新推荐文章于 2025-05-22 22:16:20 发布

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分类专栏： LeetCode@Python版文章标签： LeetCode 148 Sort List 排序列表 Python

本文链接：https://blog.youkuaiyun.com/Dby_freedom/article/details/84891854

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本文介绍了一种在O(nlogn)时间内使用常数空间复杂度对链表进行排序的方法，详细解析了归并排序在链表上的应用，通过递归将链表分为两半，然后合并两个已排序的子链表。

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

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

Reference Answer

思路分析

常见排序方法有很多，插入排序，选择排序，堆排序，快速排序，冒泡排序，归并排序，桶排序等等。。它们的时间复杂度不尽相同，而这里题目限定了时间必须为O(nlgn)，符合要求只有快速排序，归并排序，堆排序，而根据单链表的特点，最适于用归并排序。

Code

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return head
        pre, slow, fast = head, head, head
        while fast and fast.next:
            pre = slow
            slow = slow.next
            fast = fast.next.next
        pre.next = None 
        l1 = self.sortList(head)
        l2 = self.sortList(slow)
        return self.mergeLists(l1, l2)
    def mergeLists(self, l1, l2):
        if not l1:
            return l2
        if not l2:
            return l1
        root = ListNode(0)
        head = root
        while l1 and l2:
            if l1.val < l2.val:
                root.next = l1
                l1 = l1.next
            else:
                root.next = l2
                l2 = l2.next
            root = root.next
        root.next = l1 if l1 else l2
        return head.next

C++ Version

/**
 * 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) {
        if(!head || !head->next){
            return head;
        }
        ListNode *pre = head;
        ListNode *slow = head;
        ListNode *fast = head;
        while (fast && fast->next){
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
            
        }
        pre->next = NULL;
        ListNode *l1 = sortList(head);
        ListNode *l2 = sortList(slow);
        return mergeList(l1, l2);
            
    }
    ListNode* mergeList(ListNode* l1, ListNode* l2){
        if (!l1){
            return l2;
        }
        if (!l2){
            return l1;
        }
        ListNode *root = new ListNode(0);
        ListNode *head = root;
        while (l1 && l2){
            if (l1->val < l2->val){
                root->next = l1;
                l1 = l1->next;
            }
            else{
                root->next = l2;
                l2 = l2->next;
            }
            root = root->next;
        }
        root->next = (NULL != l1)? l1 : l2;
        return head->next;
    }
};