13.排序链表

算法描述

给你链表的头结点 head ，请将其按 升序 排列并返回 排序后的链表 。

原题链接

https://leetcode-cn.com/problems/sort-list/

样例

示例一：

输入：head = [4,2,1,3]
输出：[1,2,3,4]

示例二：

输入：head = [-1,5,3,4,0]
输出：[-1,0,3,4,5]

示例三：

输入：head = []
输出：[]

解法一

思路

复杂度分析,n为链表长度

时间复杂度分析：
根据二分法可知，时间复杂度为O(nlogn)
空间复杂度分析：O(logn)
对数组做归并排序的空间复杂度为O(n)，分别由新开辟数组O(n)和递归函数调用O(logn)组成，而根据链表特性：
数组额外空间：链表可以通过修改引用来更改节点顺序，无需像数组一样开辟额外空间；
递归额外空间：递归调用函数将带来O(logn)的空间复杂度，因此若希望达到O(1)空间复杂度，则不能使用递归。

代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* sortList(ListNode* head) {
        return sortList(head, nullptr);
    }
 
    ListNode* sortList(ListNode* head, ListNode* tail) {
        if (head == nullptr) {
            return head;
        }
 
        if (head->next == tail) {
            head->next = nullptr;
            return head;
        }
 
        ListNode* slow = head, *fast = head;
 
        while (fast != tail) {
            slow = slow->next;
            fast = fast->next;
 
            if (fast != tail) {
                fast = fast->next;
            }
        }
 
        ListNode* mid = slow;
        return merge(sortList(head, mid), sortList(mid, tail));
    }
 
    ListNode* merge(ListNode* head1, ListNode* head2) {
        ListNode* dummyHead = new ListNode(0);
        ListNode* temp = dummyHead;
        ListNode* temp1 = head1;
        ListNode* temp2 = head2;
 
        while (temp1 != nullptr && temp2 != nullptr) {
            if (temp1->val <= temp2->val) {
                temp->next = temp1;
                temp1 = temp1->next;
            } else {
                temp->next = temp2;
                temp2 = temp2->next;
            }
 
            temp = temp->next;
        }
 
        if (temp1 != nullptr) {
            temp->next = temp1;
        }
 
        if (temp2 != nullptr) {
            temp->next = temp2;
        }
 
        return dummyHead->next;
    }
};

解法二

思路

复杂度分析

时间复杂度分析：
O(nlogn)
空间复杂度分析：
O(1)

代码

class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null) {
            return head;
        }
        int length = 0;
        ListNode node = head;
        while (node != null) {
            length++;
            node = node.next;
        }
        ListNode dummyHead = new ListNode(0, head);
        for (int subLength = 1; subLength < length; subLength <<= 1) {
            ListNode prev = dummyHead, curr = dummyHead.next;
            while (curr != null) {
                ListNode head1 = curr;
                for (int i = 1; i < subLength && curr.next != null; i++) {
                    curr = curr.next;
                }
                ListNode head2 = curr.next;
                curr.next = null;
                curr = head2;
                for (int i = 1; i < subLength && curr != null && curr.next != null; i++) {
                    curr = curr.next;
                }
                ListNode next = null;
                if (curr != null) {
                    next = curr.next;
                    curr.next = null;
                }
                ListNode merged = merge(head1, head2);
                prev.next = merged;
                while (prev.next != null) {
                    prev = prev.next;
                }
                curr = next;
            }
        }
        return dummyHead.next;
    }
 
    public ListNode merge(ListNode head1, ListNode head2) {
        ListNode dummyHead = new ListNode(0);
        ListNode temp = dummyHead, temp1 = head1, temp2 = head2;
        while (temp1 != null && temp2 != null) {
            if (temp1.val <= temp2.val) {
                temp.next = temp1;
                temp1 = temp1.next;
            } else {
                temp.next = temp2;
                temp2 = temp2.next;
            }
            temp = temp.next;
        }
        if (temp1 != null) {
            temp.next = temp1;
        } else if (temp2 != null) {
            temp.next = temp2;
        }
        return dummyHead.next;
    }
}