题目描述

• 难点在于时空复杂度的要求
  

思路 & 代码

• 转化成：归并排序 + 合并两个有序链表 即可
• 利用快慢指针来拆分成两条链表
• 注意：链表的拆分 & 连接
• 时间复杂度O(n * logn)，空间复杂度 O(1)

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode sortList(ListNode head) {if(head == null || head.next == null){return head;}// 1. 快慢指针分出前后两部分，递归ListNode fast = head;ListNode slow = head;while(fast != null && fast.next != null){slow = slow.next;fast = fast.next.next;}// 长度为2的情况特殊考虑if(slow.next == null){slow = head;}// 拆解，递归 O(logn)fast = sortList(slow.next);slow.next = null;slow = sortList(head);// 2. 返回值，进行合并(转化成合并链表)return mergeSortedList(slow, fast);}// 链表合并O(n)ListNode mergeSortedList(ListNode head1, ListNode head2){if(head1 == null){return head2;}if(head2 == null){return head1;}if(head1.val < head2.val){head1.next = mergeSortedList(head1.next, head2);return head1;}else{head2.next = mergeSortedList(head1, head2.next);return head2;}}
}

二刷

• 好吧…记得思路是快慢指针 + 合并有序链表，但是具体咋写确实回想不起来= =
• 其实就是两个函数：快慢指针二分链表 + 合并两个有序链表，双重递归！

class Solution {// 1. 快慢指针二分链表public ListNode sortList(ListNode head) {if(head == null || head.next == null) {return head;}ListNode slow = head, fast = head;while(fast != null && fast.next != null) {slow = slow.next;fast = fast.next.next;}// 两个的情况，避免无限循环if(slow.next == null) {slow = head;}fast = sortList(slow.next);slow.next = null;slow = sortList(head);return mergeSort(slow, fast);}// 2. 合并两个有序链表 O(n)、O(1)public ListNode mergeSort(ListNode headA, ListNode headB) {if(headB == null) {return headA;}if(headA == null) {return headB;}if(headA.val < headB.val) {headA.next = mergeSort(headA.next, headB);return headA;}else {headB.next = mergeSort(headA, headB.next);return headB;}}
}