data structures and algorithms coding interview questions

It seems that the following is the most elegant implementation.Other, either node redundancy, or initialize ugly ...#!/usr/bin/env python#-*-coding:utf-8-*-classNode:def __init__(self, initdata): Self.__data=InitData self.__next=NonedefGetData (self):returnSelf.__data defGetNext (self):returnSelf.__next defSetData (Self, newdata): Self.__data=NewDatadefSetnext (Self, newnext): Self.__next=Newnextclasssincyclinkedlist:def __init__(self): Self.head=Node (None) self.head.setNext (self.head)defAdd

next time . A the defShufflepile (self): + """in the current state, the tree is adjusted so that it becomes a heap""" - #downward adjustment from "heap bottom" to "heap top", which keeps the smallest elements rising $ #This allows the heap below the I node to be the local minimum heap. $ forIinchRange ((Len (self)-2)/2,-1,-1):#N/2,..., 0 - Self.siftdown (i) - the defdeletemin (self): - """Remove Minimum element"""Wuyit = self[0]#record the

//Delete a node in a binary sort tree the PublicBstree Deletebstree (Bstree bstree,intkey) { the if(Bstree = =NULL) { - return NULL; - } the the if(Bstree.data = =key) { the //first case: leaf node the if(Bstree.left = =NULL Bstree.right = =NULL) { -Bstree =NULL; the } the //second case: node has left dial hand nodes. the if(Bstree.left! =NULL Bstree.right = =NULL) {94Bstree =Bstree.left; t

1 PackageIYou.neugle.search;2 3 Public classHash_search {4 Private Static intm = 13;5 Private Static int[] hash =New int[m];6 Private Static int[] Array =New int[] {13, 25, 2, 60, 39, 52 };7 8 Public Static voidMain (string[] args) {9 Inserthash ();TenSYSTEM.OUT.PRINTLN ("Hash table is established as follows:"); OneSystem.out.print ("["); A for(inti = 0; i ) { - System.out.print (Hash[i]); - if(I! = hash.length-1) { theSystem.out.print (","); -

　　1 //input A, B, output a+b2 /*#include 3 int main ()4 {5 int A, b;6 scanf ("%d%d", a,b);7 printf ("%d", a+b);8 }*/9 /*Ten //Enter a character to return his ASCLL code One #include A int main () - { - char A; the scanf ("%c", a); - printf ("%d", a); - }*/ - + -#include + intMain () A { at CharA; - Charb; - while(scanf ("%c%c", a,b)!=eof)//when there is a space in the middle, the input parameters should also have a space, when there is no space, you do not have to enter a space.

Hanoi Problem Solving:Hanoi (also known as Hanoi) is a puzzle toy derived from an ancient Indian legend. When big Brahma created the world, he made three diamond pillars, and stacked 64 gold discs on a pillar from bottom to top in order of size. The great Brahma commanded the Brahman to rearrange the discs from below to the other pillars in order of size. It is also stipulated that the disc cannot be enlarged on the small disc, and only one disc can be moved between the three pillars at a time.

Stack: LIFO. Stack top at the end, bottom of the stack in the front. The newly added element and the element to be deleted are saved at the end of the stack.Create a stack:functionStack () {varitems = [];/*Save the elements in the stack with an array*/ This. Push =function(e) {Items.push (e); } This. Pop =function() { returnItems.pop (); } This. Peek =function() { returnItems[length-1]; } This. IsEmpty =function() { returnItems.length = = 0; } This. Size

GCD Greatest Common divisorMethod: Euclid Algorithm (Euclidean method),"Thought" recursionIdeasCode1 Public classMain {2 Public Static intgcdintPintq) {3 if(q = 0) {4 returnp;5 }6 intr = p%Q;7 returngcd (q, R);8 }9}LCM Least common multipleThe pursuit of convenient formula method to solve.IdeasCode1 Public classMain {2 Public Static intgcdintPintq) {3 if(q = 0) {4 returnp;5 }6 intr = p%Q;7 ret

Bubble sortThe basic idea of bubble sorting is to compare two adjacent elements each time and swap them out if they are in the wrong order.If there are n numbers to sort, just n?1 the number of digits, which meansN-1 operation. The "Every trip" requires a comparison of two adjacent numbers starting from the 1th position, and the smaller oneIn the back, when the comparison is complete, move back one to continue comparing the size of the two adjacent numbers below, repeat this step until the lastN

; tointk=-1; .intJ=0; - while(JPlen-1) ,{ the //p[k] Represents a prefix, p[j] represents a suffixTenif(k==-1|| P[J]==P[k]) One { A++k; -++J; -Next[J]=k; the} -Else -{ -k=Next[k]; +} -} +}The value of Next[j] (that is, K) indicates the next move position of the J pointer when p[j]! = T[i].When J is 0 o'clock, if this time does not match, J already on the leftmost, can not move again, this time should be I pointer back move. So in the code there will be next[0] =-1; this initialization.P[K]! = P[

① problem: If there is a tuple or sequence containing n multiple elements, now you want to break it down into n separate variables. 1 L = (4, 5)2 x, y = lView Code② Advanced article:data = ['sb' ' big hammer ', (2018, 6, 2 = dataa='sb'b=' big hammer ' C=51Date= (2018, 6, 2)③data = ['sb' ' big hammer ', (2018, 6, 2 == 2018= 6= 2View CodeSo then the question comes, this is to know the list of how much of the situation to achieve, if the unknown? that

numbering unit 　　You can use the stack to do 10-in-2~9 binary operationsHere's how: a decimal number A, binary B1> will a%b, press into the stack2> Replace A with A/b3> if a is greater than 0, continue to hit 1> repeatedlyIf it is less than 0, jump out4> the elements of the stack pop up once, forming a new character (the character is the result of conversion completion)As an example:　　10 to 2 binary:10%2 = 0--into the stack--05%2 =--1 in the stack, 02%2 = 0--into the stack--0, 1, 01%2 =--1 in s

1. Code implementationdefMerge_sort (alist):ifLen (alist) : returnalist#two-part decompositionnum = Len (alist)/2 Left=Merge_sort (Alist[:num]) right=Merge_sort (alist[num:])#Merging returnmerge (Left,right)defmerge (left, right):" "merge operation, combine two ordered arrays left[] and right[] into a large ordered array" " #The subscript pointer to left and rightL, R =0, 0 result= [] whileL andrLen (right):ifLEFT[L] Right[r]: Result.append (Left[l]) L+ = 1Else: Result.append (R

1. How to find two pointsFind the fastest speed for a sorted list2. Code implementation(1) Recursive implementationdefBinary_search (Alist, item):ifLen (alist) = =0:returnFalseElse: Midpoint= Len (alist)//2ifalist[midpoint]==Item:returnTrueElse: ifitemAlist[midpoint]:returnBinary_search (Alist[:midpoint],item)Else: returnBinary_search (alist[midpoint+1:],item) testlist= [0, 1, 2, 8, 13, 17, 19, 32, 42,]Print(Binary_search (Testlist, 3))Print(Binary_search (Testlist, 13))(2) N

1. Principle:2. Code implementationdefQuick_sort (alist, Start, end):"""Quick Sort""" #Recursive exit conditions ifStart >=End:return #sets the starting element as the datum element to find the positionMID =Alist[start]#Low is the left-to-right cursor for the left of the sequenceLow =Start#High is the right-to-left cursor to the right of the sequenceHigh =End whileLow High :#If low is not coincident with high, the element with high points is not smaller than the datum element, then high

1. Basic properties of stacks and queues Stack is advanced after out; (like a bullet clip, a last-shot first) Queues are FIFO; (like waiting in line to buy ice cream, in order to take turns) Stack and queue in the implementation of the structure can have an array and a list of two forms; (1) The array structure is easy to implement;(2) The structure of the linked list is more complicated, because it involves a lot of pointer operation;1.1 Basic operation of stack structure(1) Pop op

()) { CharCH =Thestack.pop (); Output= output +ch; } returnoutput; } } classReverseapp { Public Static voidMain (string[] args)throwsIOException {String input, output; while(true) {System.out.print ("Enter A string:"); System.out.flush (); Input=getString1 (); if(Input.equals ("")) { Break; } reverser Thereverser=Newreverser (input); Output=Thereverser.dorev (); System.out.println ("Reversed:" +output); } } Public StaticString getString

value.Similarly, when an element of the RightMark position is greater than or equal to the base value, RightMark moves to the left one position to continue scanning, and the scan stops when the element of the RightMark position is less than the base value.After stopping the scan, we compare the size of the Leftmark and RightMark, if the Rightmark　　After we have put the datum values in the correct position, we see that the elements on the left side of the base value are smaller than the datum va

"Select Sort"The selection sort is based on a bubbling sort (Bubble sort) that has been improved: each visit process (pass) needs to be exchanged at most.Each visit process, to find the maximum value, when the end of the visit, the maximum value is exchanged to the correct position;Then continue to repeat the process in the remaining sublist until the n-1 visit is completed (n is the length of the list);At this point, the remaining elements in the list are automatically aligned to the correct po

= [] showing the same gender; Let len = this.list.length; while (len--) {if (This.list[len].gender = = = Gender) {Ret.push (this.list[len].name); }} return ret;} Example let people = new person ();p eople.save (' Mazey ', ' Male ');p eople.save (' John ', ' Male ');p eople.save (' Zero ', ' Male '); People.save (' July ', ' Male ');p eople.save (' Bob ', ' Male ');p eople.save (' Ada ', ' female ');p eople.save (' Cherrie ', ' female ');p eople.save (' Luna ', ' female ');p eople.save