streamfield structures

("pause");}void Push (Stack s){Elemtype e;Int J, N;if (s.stacksize = = 0){cout System ("pause");Return}if (s.top = = s.stacksize){cout System ("pause");Return}cout CIN >> N;cout for (j = 0; J {CIN >> E;* (S.base + s.top) = e; Insert a new element++s.top; Stack top pointer moves back}System ("pause");}void Pop (Stack s){Elemtype e;Int J, N;if (s.stacksize = = 0){cout System ("pause");Return}if (S.top = = 0){cout System ("pause");Return}cout CIN >> N;cout for (j = 0; J {E = * (S.base + s.top-1);

}return k;}void DFS (mgraph G, int v){Int J;cout VISITED[V] = TRUE;for (j = 0; J {if (G.kind = = DG | | G.kind = = AG)if (!visited[j] G.arcs[v][j].adj = = 1)DFS (G, J);if (G.kind = = DN | | G.kind = = an)if (!visited[j] G.arcs[v][j].adj! = INFINITY)DFS (G, J);}}void Dfstraverse (Mgraph G){int V;for (v = 0; v VISITED[V] = FALSE; Initialize to False, and change to true after traversalfor (v = 0; v if (!visited[v])DFS (G, v);} Void Bfstraverse (Mgraph g) {//The breadth-first search for the graph

again, this time based on the 10-bit value (IBID.), the allocation result (logical imagination) as shown:After the assignment ends. Next, the data in all buckets (the same principle as above) is then re-collected serially, resulting in the following data series:14 22 28 39 43 55 65 73 81 93Hill sortAn array of n=10 49, 38, 65, 97, 26, 13, 27, 49, 55, 4 for exampleFirst time gap = 10/2 = 549 38 65 97 26 13 27 49 55 41 a 1B2 a 2B3 A 3B4 a 4 b5 a 5B1A,1B,2A,2B are grouped, the numbers are the same

the left to find an element that is larger than the base element, stop, the two elements are exchanged until two pointers meet, the end of the cycle * pointer point to the The position is where the datum element should be located in the collection * eg * {8,-2, 3, 9, 0, 1, 7, 6} * Benchmark * First bm=8 * end = Lengt h-1 = 7 * start=0 * End--, We found 6:8 small, end pointer stopped, current index is j=7 * start++, until element 9 stops, current index i=3 * Exchange the elements corresp

. 3Update: The parameter is key and value, regardless of whether the key exists, it becomes the form of Key:value. Delete (not used in work) fruits{'a':'appleupdate','b':'Banana','g':'Grape','o':'Orange','Martin': -,3:5}fruits.pop (3)5fruits{'a':'appleupdate','b':'Banana','g':'Grape','o':'Orange','Martin': -} change:1, Assignment2, updatea{2:9,3:5,4:9}a[0] =8a{2:9,3:5,4:9,0:8}a[2] =8a{2:8,3:5,4:9,0:8}a.update ({3:Ten}) a{2:8,3:Ten,4:9,0:8} check a{2:8,3:Ten,4:9,0:8}a.get (2)8a[0]8two different v

, while (!thestack.isempty ()) {int Currentvertex = Thestack.peek (); int v = Getadjunvisitedvertex (Currentvertex); if (v = =-1) {Thestack.pop ();} else{vertexlist[v].wasvisited = True;thestack.push (v);d Isplayvertex (Currentvertex);d Isplayvertex (v); System.out.print ("");}} Search complete, initialize, for the next search for (int i = 0; i 5. SummaryThe graph is made up of vertices connected by edges, which can represent many real world situations, including aircraft routes, electronic circ

One, arrayArray of direct assignment declarations1, an array is stored in a plurality of content, the contents of the array is called "element";2, each element is by the key and the value Key/value key subscript$a = Array ("One" = "1111", "one" = "2222");3, is through the key to use the value4, Subscript has two types: one is an integer (indexed array), one is a string (associative array)5, "" to operate the subscript, you can also use {} Awakened interchange, the proposed use [] to specify subs

start++; } } //Return to start or end, where start and end are the datum values return end; } /** * Sort * @param a * @param start * @param end */ public static void sort (int[] A, int start, int end) { if (Start > End) { //If there is only one element, then there is no need to go down the line. return; } else{ //If there is more than one element, continue dividing both sides recursively sort down int partition = Divide (A, start, end); So

);This is not a good solution because the components set on the content pane are not visible,Typically set to this: JFrame frame=New JFrame ("Test"); JPanel p=new JPanel (); Frame.getcontentpane (). SetBackground (color.red); P.setopaque (false); // Set Transparent Frame.add (p); // Add some Components to P ... Frame.setsize (a); Frame.setvisible (true);The P.setopaque (false) above indicates that the setting Jpane is transparent to the

) Unordered array Delete (query + fill hole) O (N) Ordered array deletion (query + fill hole) O (N) Why not use data to solve everythingAn unordered array is inserted (O (1) time), and the query spends (O (N) time). An ordered array query spends (O (logn) time), but inserts a missing cost (O (N) time). And the deletion takes (O (N) time). So a data structure needs to be inserted, queried, deleted quickly, theoretically (O (1) or O (logn) time),

their definitions into a hash list; The second part lets the user enter a word, and the program gives the definition of the word.// 字典类function Dict () { this.hashTable = new HashTable(); this.save = save; this.find = find;}function save (word, description) { this.hashTable.put(word, description);}function find (word) { return this.hashTable.get(word);}// 示例let d = new Dict();d.save('Mazey', 'a strong man.');d.save('Cherrie', 'a beautiful girl.');d.save('John', 'unknown.');consol

Implementation of the Stack// 栈类function Stack () { this.dataStore = []; this.top = 0; // 栈顶位置 相当于length，不是索引。 this.push = push; this.pop = pop; this.peek = peek; this.clear = clear; this.length = length;}// push: 入栈function push (element) { this.dataStore[this.top++] = element;}// pop: 出栈function pop () { return this.dataStore[--this.top];}// peek: 取栈顶元素function peek () { return this.dataStore[this.top - 1];}// clear: 清空栈function clear () { this.top = 0;}// leng

There are three built-in data structures in Python-lists, tuple tuples, and dictionary DictItems in the list are included in square brackets , and items are separated by commasTuples and lists are very similar, except that tuples and strings are immutable , meaning that you cannot modify tuples. Tuples are defined by a comma-separated list of items in parentheses .The most common use of tuples is in the print statementAge = 22name = ' Swaroop 'Print '

"Merge Sort" Here we use recursive algorithm to keep the list in two, base case is no element in the list or only one element, because this sub-list is bound to be a positive sequence, and then gradually merge the two sorted sub-list into a new positive sequence table, until all the elements sorted."This is a process from the bottom up (bottom-up)Divides the list from the middle into two sub-lists until it reaches the bottom, with only one element in the sub-list　　Then, the two sub-lists are mer

)? {???????? This.ddata?=?ddata;???????? This.idata?=?idata,????}???? Public?void?displaylink ()? {???????? System.out.println ("{"? +?idata?+? ",?"?) +?ddata?+ "}");????}}Simple test Cases:Package?com.xingej.algorithm.datastructure.linkedlist.singlelinkedlist;import?org.junit.test;/**?*? single-linked list test? *[email protected]?erjun?2017 December 8? Morning 9:00:54?*/public?class? Linklisttest? {[email protected]???? Public?void?test ()? {???????? Linklist?list?=?new? Linklist ();???????? ?

algorithm is O (NLOGN).2: Worst of all, that's what I'm talking about. The maximum and minimum values are selected:Then look at the worst-case fast-track, when the sequence to be sorted is in a positive or reverse order, and each partition has only one sub-sequence that is less than the last one, noting that the other is empty. If the recursive tree is drawn, it is a diagonal tree. You need to perform a n‐1 recursive call at this time, and the first Division I need to go through the N‐i keyword

){ if(Array[j] key) {Array[i+ +] = Array[j];//A[i] = a[j]; i + = 1; Break; }; } for(; i ){ if(Array[i] >key) {Array[j--] =Array[i]; Break; }}} Array[i]=key; Sort (0, i); Sort (i+ 1, numsize); }} sort (0, Array.Length); returnArray;}There is also an easy-to-understand approach:Set two empty arrays left and right, traverse the entire array, and push in to leave if it encounters less critical data, otherw

The basic concept of arrays: arrays are the simplest and most commonly used data structures, but there are some caveats: (1) The distribution mode and storage location of the array; (2) initialization; (3) Advanced definition of array in different languages; (4) Multidimensional array; Array allocation methods in C + +: (1) int a[10]; Applies to situations where the length of the array is known or is not sensitive to the length of the log, such as de

Data structure: including logical structure and physical structure (logical structure in the computer storage form). Four logical structures: 1. Set structure 2. Linear structure (one-to-one) 3. Tree structure (one-to-many) 4. Graphic structure (many-to-many). Physical Structure Storage mechanism form of data element: 1. Sequential storage: The data elements are stored in the address continuous storage unit, the logical relationship and physical r

Common stored procedures for copying table structures --An important attribute of the Transfer object --1. Property Property name Type description------------------------------------------------------------------------CopyAllDefaults Boolean All default valuesCopyAllObjects Boolean All objectsCopyAllRules Boolean All RulesCopyAllStoredProcedures Boolean All stored proceduresCopyAllTables Boolean All TablesCopyAllTriggers Boolean all triggersCopyAllUse