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from its previous element, execute a[count]=a[i],count++;Question 16: How to find the second largest number in an array.Method 1: Sort all the elements and the second largest number in the second positionMethod 2: Use two variables to record the first and second largest number, for the maximum initial value of the primary element, the second largest number is initialized to the smallest negative numbers. Each time the current element is compared with the maximum value, if it is greater than the
var arraynumber: [Int] = [2,4, 6, 7, 3, 8, 1]Bubble sortfunc Maopao (var array: [int]), [int] { for var i = 0; i count; i++ { for var j = I;j count; j + + {if array[j] > array[j +1] {var temp = Array[j]ARRAY[J] = array[j+1]array[j+1] = Temp}}}return Array} Let array2 =maopao([2,4, 6, 7, 3, 8, 9 , 5])Cond... Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. Swift sorting algorithms and
Huashan Big BroSort by: Topological sorting algorithmDictionary ordering algorithmProgramming Zhu Ji Nanxiong: Ordering by bitmapTree: Red and Black tree summaryB + Tree and b* Tree SummaryB-Tree Summary SummaryBalanced binary tree (AVL tree) Summarytrie--Dictionary TreeGraph traversal: Depth-first traversal and breadth-first traversalMinimum spanning tree: minimum spanning tree-prim algorithm and Kruskal algorithmShortest path: Shortest path-dijkstra algorithm and Floyd algorithmCommon classica
, vi belongs to V-S. Then, the shortest path is compared with the source point A to the path length of the node vi, the minimum length is the shortest path from source point A to Node VI. Finally, the end of the shortest path (i.e. VI) is added to the S(3) Repeat step (2) until the shortest path of all endpoints is found, that is, the number of nodes in S is equal to the number of nodes in V.The time complexity of the Dijkstra algorithm is O (n^2), the spatial complexity depends on the storage m
Data structures and algorithms (C # implementation) series-demonstration (II)
Heavenkiller (original)
Public static void ShowGeneralTree_travel ()
{
IEnumerator tmpIEnum;
Tree. TraversalType travelType = 0;
// --------------------- Prompt ----------------------------
Console. WriteLine ("please choose a No. of a item you want to travel :");
Console. WriteLine ("
Data structures and algorithms (C # implementation) series --- tree (2)
Heavenkiller (original)
Public class InOrder: IPrePostVisitor
{
Private IVisitor visitor;
Public InOrder (IVisitor _ vis) {visitor = _ vis ;}
# Region IPrePostVisitor Member
Public void PreVisit (object _ obj)
{
// TODO: Add InOrder. PreVisit implementation
}
Public void Visit (object _ ob
Data structures and algorithms (C # implementation) series --- generalized tree (II)
Heavenkiller (original)
Public override object Key {get {return this. key ;}}
Public override uint Degree {get {return this. degree ;}}
// Public override uint Height {get {return this. height ;}}
Public override bool IsEmpty () // property takes the place of IsEmpty ()
{Return
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 //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
only inherit one parent classPolymorphicFour, algorithm effectiveness analysis:4.1 Time Complexity: Https://baike.baidu.com/item/%E6%97%B6%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6/1894057?fr=aladdin (Baidu Connection) F (N) =n! (factorial) n-times with a time complexity of n4.2 Space Complexity: Https://baike.baidu.com/item/%E7%A9%BA%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6/9664257?fr=aladdin (Baidu Connection)V. Abstract classTo include one or more abstract methods (without a method body): To implement
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
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