binary-search

问题 I have some data that is stored in a sorted vector. This vector is sorted by some key. I know the STL has an algorithm for checking if an element is in this sorted list. This means I can write something like this: struct MyData { int key; OtherData data; }; struct MyComparator { bool operator()( const MyData & d1, const MyData & d2 ) const { return d1.key < d2.key; } }; bool isKeyInVector( int key, const std::vector<MyData> &v ) { MyData thingToSearchFor; thingToSearchFor.key = key; return

问题 Okay so this is for a school assignment. I have had no problems doing a recursive binary search but the assignment specifically says that the method should only take 2 arguments, the list, and the item you are searching for. This is where I am getting a little lost. public int binarySearch(List<Card> cards, Card key) { int mid = (cards.size()) / 2; if(cards.size() == 1) { if(key.equals(cards.get(0))) { return 0; } } else { if(key.equals(cards.get(mid))) { return mid; } else if(key.compareTo

问题 I am watching the Berkley Uni online lecture and stuck on the below. Problem : Assume you have a collection of CD that is already sorted. You want to find the list of CD with whose title starts with "Best Of." Solution : We will use binary search to find the first case of "Best Of" and then we print until the tile is no longer "Best Of" Additional question : Find the complexity of this Algorithm. Upper Bound : Binary Search Upper Bound is O(log n), so once we have found it then we print let

问题 I am watching the Berkley Uni online lecture and stuck on the below. Problem : Assume you have a collection of CD that is already sorted. You want to find the list of CD with whose title starts with "Best Of." Solution : We will use binary search to find the first case of "Best Of" and then we print until the tile is no longer "Best Of" Additional question : Find the complexity of this Algorithm. Upper Bound : Binary Search Upper Bound is O(log n), so once we have found it then we print let

问题 Given a sorted list of numbers, I need to find the smallest number that is greater than a given number. Consider this list: arr=[1,2,3,5,7,11,101,131,151,181,191,313,353,373,383] Say the specified number is 320. Then, my method should return 353 as 353 is the smallest number greater than 320. I am trying to use a slightly modified form of binary search; however on execution the program goes into infinite loop. def modBinarySearch(arr,x): l=len(arr) mid=l/2 if arr[mid]>=x and arr[mid-1]<x:

问题 I'm just starting to learn about parallel programming, and I'm looking at binary search. This can't really be optimized by throwing more processors at it right? I know it's supposedly dividing and conquering, but you're really "decreasing and conquering" (from Wikipedia). Or could you possibly parallelize the comparisons? (if X is less than array[mid] , search from low to mid - 1 ; else if X is greater than array[mid] search from mid + 1 to high , else return mid , the index of X ) Or how

问题 I need help writing a program that uses binary search to recursively compute a square root (rounded down to the nearest integer) of an input non-negative integer. This is what I have so far: import java.util.Scanner; public class Sqrt { public static void main(String[] args) { Scanner console = new Scanner(System.in); System.out.print("Enter A Valid Integer: "); int value = console.nextInt(); calculateSquareRoot(value); } public static int calculateSquareRoot(int value) { while (value > 0) {

问题 Is there a function in that uses binary search, like lower_bound but that returns the last item less-than-or-equal-to according to a given predicate? lower_bound is defined to: Finds the position of the first element in an ordered range that has a value greater than or equivalent to a specified value, where the ordering criterion may be specified by a binary predicate. and upper_bound : Finds the position of the first element in an ordered range that has a value that is greater than a

问题 I am not able to solve the following problem optimally nor finding an approach to do this anywhere. Given a N × M matrix in which each row is sorted, find the overall median of the matrix. Assume N*M is odd. For example, Matrix = [1, 3, 5] [2, 6, 9] [3, 6, 9] A = [1, 2, 3, 3, 5, 6, 6, 9, 9] Median is 5. So, we return 5. Note: No extra memory is allowed. Any help will be appreciated. 回答1: Consider the following process. If we consider the N*M matrix as 1-D array then the median is the element

问题 What is the difference between binary search and binary search tree? Are they the same? Reading the internet it seems the second is only for trees (up to 2 children nodes) and binary search doesn't follow this rule. I didn't quite get it. 回答1: Binary Search Trees A node in a binary tree is a data structure that has an element, and a reference to two other binary trees, typically called the left and right subtrees. I.e., a node presents an interface like this: Node: element (an element of some