bisearch

/////------------------------------------------------------------------------------namespacetestapp{usingSystem; usingSystem.Collections.Generic; usingSystem.Text.RegularExpressions; classRunner {Static voidMain (string[] args) { int[] arr = {3,5,6,7,8,9, -, at, $, About, the, the, the, the, the, the, the, the,98,534,555,676,878,988,1365 }; intNumber =555; intresult =Search (arr, number); Console.WriteLine (result); Console.readkey (); } Public Static intSear

caller to inform the index range of the data source, and provides the logic for comparing the data at the specified index (as a delegate ). In this way, algorithms only need to operate on indexes and are isolated from specific data sources. The comparison logic and can be customized by users. Example: Code Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/--> Searchalgorithm. bisearchmatch match = Delegate ( Int Index){ Ret

binary tree. The first half is divided into the left subtree, and the second half is the right subtree. The number of searches by the half lookup method is exactly the number of layers where the value is located. In the case of equal probability Log2 (n + 1)-1 The Code is as follows: // Data is the array to be searched, x is the value of the Data to be searched, beg is the start of the search range, and last is the end of the search range// Non-Recursive MethodInt

the number of layers where the value is located. In the case of equal probability, aboutLOG2 (n+1)-1[CPP]View Plaincopy Data is the array to find, X is the value to look for, beg is the start of the lookup range, and last is the end of the lookup range Non-recursive method int Bisearch (int data[], const int x, int beg, int last) { int mid; //Middle position if (Beg > last) { return-1; } While (Beg { Mid = (b

method described in O (NlogN) algorithm 2 needs to traverse all dp [0] ~ When solving dp [I]. Dp [I-1], so can reduce each traversal? In fact, when I is a little larger, j =, 2, and so on may not be able to meet the requirements of dp [j]> dp [I]-1, and it is useless to traverse them, after finding a j that meets both of the preceding conditions, we can determine that items smaller than dp [j] do not need to be considered. Therefore, we need to find items that meet val [j].

number of layers in which the value is located. In the case of equal probability, about LOG2 (n+1)-1 Copy Code code as follows: The data is the array to look for, X is the value to look for, beg is the start of the lookup range, and last is the lookup range terminated Non-recursive method int bisearch (int data[], const int x, int beg, int last) { int mid;//Middle Position if (Beg > last) { return-1; } while (Beg { Mid = (beg

, min; Int[] arr={3,2,1,6,9,5}; The initial value of Max=min=arr[0];//max and Min can be an array of any two values, but the following loop must start with subscript 0 (i=0) for (int i=1;iOperation Result:Binary (two-point search): Example: Class Bisearch {/* binary (binary) lookup: algorithm idea: requires sequence must be ordered (ascending/descending), because only in order to determine the scope of the data, through constant binary

binary tree. the first half is divided into the left subtree, and the second half is the right subtree. The number of searches by the half lookup method is exactly the number of layers where the value is located. In the case of equal probability Log2 (n + 1)-1 The code is as follows: // Data is the array to be searched, x is the value of the Data to be searched, beg is the start of the search range, and last is the end of the search range// Non-recursive methodInt

Static voidMain (string[] args) { Char[] arr =New[] {'a','b','C','D','e'}; Charv ='D'; varindex = Bisearch (arr,0,4, V); Console.WriteLine (index); Char[] arr2 =New[] {'b','C','D','e','F' }; CharV2 ='a'; varIndex2 = Bisearch (ARR2,0,4, v2); Console.WriteLine (INDEX2); Console.read (); } Static intBisearch (Char[] arr,intBintECharv) {intMinindex = B, Maxindex =e

subscript ordinal of the object, and returns-1 on failure. intBISEARCH2 (intR[],intLowintHighintk) {if(low>High )return-1; Else { intMid= (Low+high)/2; if(r[mid]==k)returnmid; Else if(r[mid]k)returnBISEARCH2 (r,mid+1, high,k); Else returnBISEARCH2 (r,low,mid-1, K); } } Public Static voidMain (string[] args) {bisearch bs=NewBisearch (); intr[]={1,2,3,4,5}; SYSTEM.OUT.PRINTLN (BS

divided into the left subtree, and the second half is the right subtree. The number of searches by the half lookup method is exactly the number of layers where the value is located. In the case of equal probability Log2 (n + 1)-1 Copy codeThe Code is as follows:// Data is the array to be searched, x is the value of the Data to be searched, beg is the start of the search range, and last is the end of the search range// Non-Recursive MethodInt BiSearch

Problem:Find an ordinal (dictionary-ordered) string array of elements in arr that are equal to the string V, and if more than one element satisfies this condition, returns the largest of the ordinal.int Bisearch (char** arr, int b, int e, char* v){There are two scenarios for the end of the loop:If the minindex is an even number minindex==maxindex;otherwise it is minindex==maxindex-1;while (minindex{Midindex = Minindex + (maxindex-minindex)/2;//not Use

1. BackgroundFor example, an integer x is a set of numbers in a sequence of columns sorted by size, and we want to find the index position of x in the sequence.For example, the sequence is arranged from small to large:-3,-2,0,4,5,7,12,64We want to find the position of the number 7, if it is a linear lookup, the time complexity is O (n), if you find with binary, the time complexity is O (log (n)), because each time the binary, the calculation of less than half, so take the logarithm.2. CodePackag