leetcode链表--3、sort-list(单链表排序) 题目描述 Sort a linked list in O(n log n) time using constant space complexity. 思路: 冒泡排序、选择排序、插入排序,时间复杂度为O(n^2); 快速排序、归并排序、堆排序,时间复杂度为O(nlogn); 基数排序、计数排序,时间复杂度都是O(n)。 首先确定时间复杂...
java代码: 1/**2* Definition for singly-linked list.3* class ListNode {4* int val;5* ListNode next;6* ListNode(int x) {7* val = x;8* next = null;9* }10* }11*/12publicclassSolution {13publicListNode sortList(ListNode head) {14if(head==null||head.next==null)returnhead;15ListN...
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 解题思路:回想排序算法,线性对数时间复杂度的有:快速排序,堆排序,归并排序,前两种暂时没实现,...
时间复杂度:O(n^2) 稳定排序 (a)严格按照定义来说,将未排序部分的元素插入时,需从后往前扫描,然后将大于元素值(假设是从小到大排序)依次后移,找到插入位置后跳出循环,插入元素。 (b)由于这里是对单链表排序,故无法从后往前扫描,另外链表这种结构决定了没必要进行依次后移这种操作,操作指针即可。严格按照定义的...
【Leetcode】Insertion Sort List https://leetcode.com/problems/insertion-sort-list/ 题目: Sort a linked list using insertion sort. 思路: 头插法。用头结点可以简化插入链表时候的操作,因为要考虑插入链表中间和表头两种情况,插入表头时,head就要更新,还要判断pre指针是否为空...
Sort a linked list in O(n log n) time using constant space complexity. 解题思路:题目限定了时间必须为O...
Leetcode 147. Insertion Sort List LinkedList Sort a linked list using insertion sort. 题目大意:使用插入排序对一个链表排序。 思路:可以构建一个临时的链表,然后将待排序的链表的每一个节点插入到临时链表中 1/**2* Definition for singly-linked list.3* struct ListNode {4* int val;5* struct ...
| 1 | A | f | 2500 | | 2 | B | m | 1500 | | 3 | C | f | 5500 | | 4 | D | m | 500 | LeetCode Question Find Minimum in Rotated Sorted Array Deion: Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.(i.e., 0 1 2 4 ...
LeetCode – Sort List: Sort a linked list in O(n log n) time using constant space complexity. Analysis If this problem does not have the constant space limitation, we can easily sort using a sorting method from Java SDK. With the constant space limitation, we need to do some pointer ma...
[LeetCode] Insertion Sort List 简介:Sort a linked list using insertion sort.解题思路 对于得到结点current的插入位置,从头结点开始遍历,直到遍历到值大于等于节点current的结点,然后将从该结点到current的前驱结点的所有结点的值依次和current结点的值交换,从而达到将该节点插入所遍历到的位置的目的。实现代码/***...