代码如下: package com.wangzhu.arrays; import java.util.Arrays; import java.util.Collections; public class ArraysDemo { /** * @param args */ public static void main(String[] args) { Dog[] dogs = new Dog[] { new Dog(5), new Dog(2), new Dog(19), new Dog(21)
按sort排序,sort为空的在后面 select * from 表名 order by (case when sort is null or sort='' then 1 else 0 end),sort select t.*, t.rowid from hs_tr_goods_attach_rel t order by (case when t.goods_order is null or t.goods_order = '' then
题目 According to Wikipedia: Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted
Sort List Sort a linked list in O(n log n) time using constant space complexity. Have you been asked this question in an interview? Yes 说明:归并排序: 时间 O(nlogn),空间 O(1). 每次将链表一分为二, 然后再合并.快排(用两个指针) /** * D
Sort a linked list using insertion sort. 利用插入排序对一个链表进行排序 思路和数组中的插入排序一样,不过每次都要从链表头部找一个合适的位置,而不是像数组一样可以从要插入的位置开始从后往前找合适的位置 class Solution(object): def insertionSortList(self, head): dummy = ListNode(-1) dummy.next,cur= head,head while cur and cur.next:
Sort a linked list in O(n log n) time using constant space complexity. Hide Tags Linked List Sort 基于单项链表的排序,时间为nlogn ,O(1)空间,其实及将数组的快速排序用链表实现,并用递归来维护拆分与合并. #include <iostream> using namespace std; /** * Definition for singly-linked list. */ struct
Bubble sort Basic Method: import random nums = [random.randint(1,20) for _ in range(10)] #制作一个无序数字列表 print(nums) length = len(nums) for i in range(length - 1): for j in range(length - 1 - i): if nums[j] > nums[j + 1]: nums[j], nums[j + 1] = nums[j +