持之以恒

给一个字符串，求这个字符串的子集中中最长的回文串。返回这个回文串的长度。思路：求子集+全排列。

【代码】裁剪地毯java。

【代码】答编程题如果需要自己写输入，如何写io。

一些细节处理错误导致程序错误。

子集

二叉搜索树中的插入操作非递归class Solution { public TreeNode insertIntoBST(TreeNode root, int val) { if (root == null) { return new TreeNode(val); } TreeNode pos = root; while (pos != null) { if (val < pos

微观层面

剑指offer中的一道题目。具体题目百度一下即可。自己写了一个双指针法。有漏洞。# class ListNode:# def __init__(self, x):# self.val = x# self.next = None## # @param pHead1 ListNode类 # @param pHead2 ListNode类 # @return ListNode类class Solution: def FindFirstComm

# -*- coding:utf-8 -*-# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution: # 返回ListNode def ReverseList(self, pHead): # write code here pre = None cur = pHead .

https://leetcode-cn.com/problems/xuan-zhuan-shu-zu-de-zui-xiao-shu-zi-lcof/solution/xuan-zhuan-shu-zu-de-zui-xiao-shu-zi-by-leetcode-s/

文章目录前序中序后序前序def preorder(tree): ''' :param tree: :return: des:根左右 ''' if tree == None: return print(tree.getRootVal()) preorder(tree.getLeftChild()) preorder(tree.getRightChild())中序def inorder(tree): if

class BinaryTree: def __init__(self, rootObj): self.key = rootObj self.leftChild = None self.rightChild = None def insertLeft(self, newNode): if self.leftChild == None: self.leftChild = BinaryTree(newNod

文章目录嵌套列表实现链表实现嵌套列表实现def binary_tree(r): return[r, [], []]def insert_left(root, new_branch): t = root.pop(1) if len(t) > 1: root.insert(1, [new_branch, t, []]) else: root.insert(1, [new_branch, [], []]) return rootdef insert_right(root, ne

# -*- encoding: utf-8 -*-"""@File : 123.py@Time : 2021-10-07 18:11@Author : XD@Email : gudianpai@qq.com@Software: PyCharm"""#实现ADT Map （字典）class HashTable: def __init__(self): self.size = 11 # 可以任意设置，但为了便于求解，应该设为素数 se

文章目录冒泡排序选择排序插入排序谢尔排序冒泡排序# -*- encoding: utf-8 -*-"""@File : demo.py@Time : 2021-09-22 21:29@Author : XD@Email : gudianpai@qq.com@Software: PyCharm"""def bubble_sort(l): for i in range(len(l) - 1): for j in range(len(l) - i -

https://zhuanlan.zhihu.com/p/145366243这个代码写的比较详细。。https://zhuanlan.zhihu.com/p/348134568

关于动态规划的一些理解https://www.zhihu.com/question/39948290官方解释https://leetcode-cn.com/problems/coin-change-2/solution/ling-qian-dui-huan-ii-by-leetcode-soluti-f7uh/使用二维数组解决问题：多少种找零钱方法https://blog.youkuaiyun.com/zouyutu5296/article/details/88657963?ops_request_mis

https://blog.youkuaiyun.com/nyz5211314/article/details/105990555

201.txtimport turtleimport randomdef get_maze_info(filename): with open(filename, 'r') as f: fl = f.readlines() maze_list = [] for line in fl: line = line.strip()#去掉 /n line_list = line.split(" ") maze_list.append(line_list) return maze.

# -*- encoding: utf-8 -*-"""@File : datastruck.py@Time : 2021-07-04 18:19@Author : XD@Email : gudianpai@qq.com@Software: PyCharm"""def movedisk(diskindex, fromtower, targettower): print("Move disk {} from {} to {}".format(diskindex, fr

# -*- encoding: utf-8 -*-"""@File : datastruct.py@Time : 2021-07-02 17:09@Author : XD@Email : gudianpai@qq.com@Software: PyCharm"""import turtlet = turtle.Turtle()def get_midpoint(a, b): ax, ay = a bx, by = b return ((ax +

import turtlet = turtle.Turtle()t.pencolor('red')t.pensize(3)for i in range(5): t.forward(300) t.right(144)t.hideturtle()turtle.done()import turtlet = turtle.Turtle()t.pencolor('red')def drawSpiral(t, lineLen): if lineLen > 0:

def listsum(numList): if len(numList) == 1: return numList[0] else: return numList[0] + listsum(numList[1:])print(listsum([1,3,5,7,9]))25

class Node: def __init__(self, initdata): self.data = initdata self.next = None def getData(self): return self.data def getNext(self): return self.next def setData(self, newdata): self.data = newda

import osimport randomimport stringdef mkdir(): if not os.path.exists("img"): os.mkdir("img") print("creat img successfully!") else: print("img file exist!")def gen_code(): list = random.sample(string.ascii_letters +

#node类实现#https://www.jianshu.com/p/7964c30197c0class Node(): def __init__(self, initdata): self.data = initdata self.next = None def getData(self): return self.data def getNext(self): return self.next de

#默认右端为队首class Deque: def __init__(self): self.items = [] def isEmpty(self): return self.items == [] def addFront(self, item): self.items.append(item) def addRear(self, item): self.items.insert(0, item) d

import randomfrom pythonds.basic.queue import Queueclass Printer: def __init__(self, ppm): self.pagerate = ppm self.currentTask = None self.timeRemaining = 0 def tick(self): if self.currentTask != None:

from pythonds.basic.queue import Queuedef hotPota(nameList, num): simqueue = Queue() for name in nameList: simqueue.enqueue(name) print("Queue init:", simqueue.items) while simqueue.size() > 1: for i in range(num):

from pythonds.basic.stack import Stackdef dicideBy2(num): s = Stack() while num > 0:#此处是 > 0 还是 == 需要思考 rem = num % 2 s.push(rem) num = num // 2 binString = [] while not s.isEmpty(): binString.ap

中缀转后缀from pythonds.basic.stack import Stackfrom string import *def infixToPostfix(infixexpr): # 这里创建一个字典是为了后面 优先级 的比较 prec = {} prec["*"] = 3 prec["/"] = 3 prec["+"] = 2 prec["-"] = 2 prec["("] = 1 # 实例化 opstack =

如果只有圆括号from pythonds.basic.stack import Stack#圆括号匹配()def parChecker(S): stack = Stack() for i in range(len(S)): if S[i] == '(': stack.push(S[i]) if S[i] == ')': stack.pop() if stack.isEmpty() == True:

http://gis4g.pku.edu.cn/course/pythonds/

import timedef sum(n): start = time.time() sum = 0 for i in range(1, n + 1): sum += i end = time.time() return sum, end - startfor i in range(100): print("sum is:{}, requried {:.10f} seconds".format(sum(100000)[0],sum(1000

5是手里的牌，7是拿出来准备插入的牌…然后把四拿出来…把四放在一个变量存起来，然后5,7往右边挪…四放进来然后七往右边挪一下…依次进行…写代码怎么挪我要摸多少次牌？n-1次插入排序，复杂度O(n平方)初始时手里(有序区)只有一张牌每次(从无序区)摸一张牌，插入到手里已有牌的正确位置示例代码def insert_sort(li): for i in range(1,len(li)):#i表示摸到的牌的下标 tmp = li[i] j =.

选择排序先遍历一遍找到最小的，然后再遍历一遍找到第二个最小的…先写一个最简单版的def select_sort_simple(li): li_new = [] for i in range(len(li)): min_val = min(li) li_new.append(min_val) li.remove(min_val) return li_new测试一下li = [3,4,5,6,1,7,2,8,9]print(

首先简述一下排序排序：将一组无序的记录序列调整为有序的记录序列列表排序：将无序列表变为有序列表输入：列表输出：有序列表升序与降序内置的排序函数：sort()常见的排序算法lowB三人组冒泡排序，选择排序，插入排序niuB三人组快速排序，堆排序，归并排序others:希尔排序，计数排序，基数排序下图是一个冒泡排序的过程冒泡排序冒泡排序(buuule sout)列表中每两个相邻的数，如果前面的比后面的大，则交换这两个数一趟排序完成后，则无序区减少一个数，有序区增加一个数整

顺序查找：从头走到尾时间复杂度O(n)代码示例def linear_search(l,val): for i, v in enumerate(list): # print(i) # print(v) if v == val: return i else: return -1测试l = [1,2,3,4,5,6,7]val = 3result = linear_search(l,val)p

时间复杂度：代码示例def binary_serch(l,val): left = 0 right = len(l)-1 while left <= right: mid = (left + right) // 2 if l[mid] == val: return mid elif l[mid] > val: right = mid -1 else: