data structures and algorithms coding interview questions

(e e);//replace element in collection * "9" void Add (e e);//At the previous position of the current index or the current Adds a new element to the collection at the back of the index position. *//whether or not to add new elements before or after the current index position depends on whether you are traversing sequentially or in reverse order * If it is {@link#next}, insert at the previous position of the current position *//if it is {@link#previous} to inser

The maximum value in the output array A and its subscript#include Data structures and algorithms-maximum values in a string output array

Title: Define the Fibonacci sequence as follows:/0 N=0F (n) = 1 n=1\ f (n-1) +f (n-2) n=2Enter N to find the nth item of the sequence in the quickest way.#include Data structures and algorithms-string Fibonacci to find the nth item

(self) self.engine.accept (visitor) self.body.accept (Visito R) forWheelinchself.wheels:wheel.accept (visitor)#This is our visitor, and every change is here .classPrintvisitor:defVisitwheel (self, Wheel):Print "Visiting"+wheel.name+"Wheel" defVisitengine (self, Engine):Print "Visiting engine" defvisitbody (self, body):Print "Visiting Body" defVisitcar (self, car):Print "Visiting Car"if __name__=='__main__': Car=Car () visitor=printvisitor () car.accept (visitor)Learn some Python basics

Package Com.js.ai.modules.pointwall.testxfz;class ordarray{private long[] a;private int nelems;public OrdArray (int max ) {a=new long[max];nelems=0;} public int size () {return nelems;} Insert method public void Insert (Long value) {int j;for (j=0;j　　Data structures and algorithms: the dichotomy Demo

True Second, collections. Ordereddict: Dictionary with OrderOrdereddict internally maintains a doubly linked list to record the order in which key values are inserted, update the key value does not affect the original order, insert key value is inserted at the end, so its memory consumption is twice times the normal dictionary ,From collections Import ORDEREDDICTD = Ordereddict () d[' a '] = 1d[' c '] = 3d[' b '] = 2print (d, D.items ()) Ordereddict (' A ', 1), (' C ', 3)

-set receives a tree height effect and is less efficient. Therefore, there are two kinds of improved heuristic algorithms.The first is to merge by rank, and for each node to record his height, change the father of the smaller node to the larger one. If both sides are equal, select one as the father, and the rank of the root node plus one.The second is path compression, and each time the find-set is called, the father of the node found on the path is changed directly to the root node.1make-SET (x

1. The principle of hill sorting2. Code implementationdefShell_sort (alist): N=Len (alist)#Initial StepGap = N/2 whileGap >0:#Insert Sort by step forIinchRange (Gap, N): J=I#Insert Sort whileJ>=gap andALIST[J-GAP] >Alist[j]: alist[j-GAP], alist[j] = alist[j], alist[j-Gap] J-=Gap#get a new step sizeGap = GAP/2alist= [54,26,93,17,77,31,44,55,20]shell_sort (alist)Print(alist)Attention:(1) Hill sort uses the idea of splitting up and merging, splitting apart3. Complexity of TimeOp

node is now Point to the head of the now node's next pointer; Set the now node to the head of the new rollover completion node header now; Point to head now, and so on, the next pointer to the previous node of the now node. 3. Two points search common scenes Finding a number in an ordered sequence; For example, given an array of arr, determine if the integer m is in arr (Train of thought: Determine the size of the mid-to-m re

; J ) {System.out.print (A[j]+ " "); } System.out.println (""); } //Bubble Sort Public voidBubblesort () {intOut , in; for(out = nElems-1, out > 1; out--) { for(in = 0; in ) { if(A[in] > a[in + 1]) {swap (in,+ 1); } } } } //Insert Sort Public voidInsertionsort () {intin, out; //Out are dividing line for(out = 1; out ) { //Remove marked item Longtemp =A[out]; //start shifts at outin =Out ; wh

The implementation of the dictionary// 字典类function Dictionary () { this.add = add; this.dataStore = new Array(); this.find = find; this.remove = remove; this.showAll = showAll; this.count = count; this.clear = clear;}function add (key, value) { this.dataStore[key] = value;}function find (key) { return this.dataStore[key];}function remove (key) { delete this.dataStore[key];}function showAll () { let datakeys = Array.prototype.slice.call(Object.keys(this.dataStore)

node1.next = node3; node3.previous = node1; node3.next = node2; node2.previous = node3; node2.next = node4; node4.previous = node2; }};DList.prototype.advance = function (n, item) { let currNode = this.find(item); while (n--) { this.goPrevious(currNode); }};// 示例let names = new DList();names.insert('Mazey', 'head');names.insert('Cherrie', 'Mazey');names.insert('John', 'Cherrie');names.insert('Luna', 'John');names.insert('Ada', 'Luna'

efficiency: In a merge sort, the maximum value of this secondary storage space is not more than N, so the spatial complexity of the merge algorithm is θ (n) in order to combine the sub-sequences to use additional storage space.Time efficiency: The merging algorithm is a typical divide-and-conquer algorithm with a time complexity of 0 (n log n).　3. Algorithm examplesMergesort.javaPackage Com.test.sort.merge;public class MergeSort {/** * @param args */public static void main (string[] args) {//TO

" definition of algorithmic time complexity"At the time of the algorithm analysis, the total number of executions of the statement T (N) is a function of the problem size n, which then analyzes the change of T (n) with N and determines the order of magnitude of T (N). The time complexity of the algorithm, which is the time measurement of the algorithm, is recorded as: T (n) = O (f (n)). It indicates that with the increase of the problem size n, the growth rate of the algorithm execution time is

]; while (end>start) { while (end>startarrys[key)//If there is nothing smaller than the key value, compare next until there is a smaller swap position than the key value, And then compare it back to the past End--; if (arrys[end]) { arrys[end]; arrys[arrys[start]; arrys[tmp; } while (end>startarrys[start]) Start+ +; if (arrys[key) {

Bucket sortBarrel sequencing has been used since the 1956, and the basic idea of the algorithm is proposed by E.J.ISSAC and R.c.singleton.This algorithm is like having 11 barrels, numbered from 0~10. Each occurrence of a number, in the corresponding number of buckets put aA little flag, and then just count the few small flags in each bucket OK. For example, there are 1 small flags in bucket 2nd, indicating2 appeared once, and in the 3rd barrels there were 1 small flags, indicating that 3 appeare

can put in *, in Put +, that is abc*+.A * (b+c) turn suffix:Read a, read *, later found (can not be placed *, (B+C * Priority High, read, followed by), can be changed to bc+, and a * together, that is abc+*.This is a queue, but the structure to use is the stack, because it comes in. The arithmetic here does not have to be pressed into the stack, directly output.a+b-c Turn suffix:Direct output A, read +, stack empty press into the stack, then output B, stack non-empty, pop +, read to-,+ priority

and J to point to the leftmost and rightmost of the sequence, respectively. At first, the variable I points to the leftmost part of the sequence, which is the number 6, and J points to the far right of the sequence, pointing to the number 8, as shown inFirst J moves from right to left, so long as it finds a number smaller than 6, it stops, and J is in the 5 position. I then start from left to right, as long as I find a number greater than 6 to stop, so it to the position of 7This is where the t

Using system;using system.collections.generic;using system.linq;using system.text;using System.Threading.Tasks; Namespace selectsort{class Program {static void Main (string[] args) {int[] Arry = {1, 4, 5,6,2,9,0,8,3,10}; Selectesort (Arry, Arry. Length); for (int i = 0; i Because of the two for loops, the time complexity is O (n^2), which is a simple but very slow sort.Selection and sequencing of algorithms and

(String)); }Second, the use of the stack-delimiter matching.public int charat (char c) {for (int i = 0; i will ({Press into the stack once, once encountered)}] will be compared with the popup element, if it matches, then match. If not)}], the left sign of the stack pops up and the hint is where the specific right symbol type is missing. This is a tool that can be implemented using stacks.Stack of data into the stack and the time complexity of the