binary-search

问题 I am stuck up with two time complexities. To do a binary search with sorted array is O(logN). So to search an unsorted array we have to sort it first so that becomes O(NlogN). So then we can perform binary search which gives the complexity as O(N) but I have read that it could be O(NlogN). Which is correct? 回答1: Binary Search is for "Sorted" lists. The complexity is O(logn). Binary Search does not work for "un-Sorted" lists. For these lists just do a straight search starting from the first

问题 I'm trying to write a function in Python that finds the first number in a sorted list greater than a specific value that I pass in as an argument. I've found examples online that use simple list comprehensions to achieve this, but for my purposes I need to be performing this operation frequently and on large lists, so a search that runs in linear time is too expensive. I've had a crack at writing an iterative binary search-like function to achieve this, though I'm coming across some edge

问题 I am trying to implement a program that will take a users input, split that string into tokens, and then search a dictionary for the words in that string. My goal for the parsed string is to have every single token be an English word. For Example: Input: aman Split Method: a man a m an a m a n am an am a n ama n Desired Output: a man I currently have this code which does everything up until the desired output part: import java.util.Scanner; import java.io.*; public class Words { public static

问题 I've heard that it's possible to implement binary search over a doubly-linked list in O(n) time. Accessing a random element of a doubly-linked list takes O(n) time, and binary search accesses O(log n) different elements, so shouldn't the runtime be O(n log n) instead? 回答1: It's technically correct to say that the runtime of binary search on a doubly-linked list is O(n log n), but that's not a tight upper bound. Using a slightly better implementation of binary search and a more clever analysis

问题 Given an array of integers, find the local minima. An element A[i] is defined as a local minimum if A[i-1] > A[i] and A[i] < A[i+1] where i = 1...n-2. In case of boundary elements, the number has to be just smaller than its adjacent number. I know if there is only one local minimum, then we can solve with modified binary search. But if it is known that there exist multiple local minima in the array, can it be solved in O(log n) time? 回答1: If the array elements are not guaranteed to be

问题 In a general binary search, we are looking for a value which appears in the array. Sometimes, however, we need to find the first element which is either greater or less than a target. Here is my ugly, incomplete solution: // Assume all elements are positive, i.e., greater than zero int bs (int[] a, int t) { int s = 0, e = a.length; int firstlarge = 1 << 30; int firstlargeindex = -1; while (s < e) { int m = (s + e) / 2; if (a[m] > t) { // how can I know a[m] is the first larger than if(a[m] <

问题 How to solve the problem with duplication in Binary Search Tree? 回答1: I am not really sure what you are asking. But that won't stop me from posting an answer. Usually, duplicate keys are disallowed in a BST. That tends to make things a lot easier, and it is a condition that is easy to avoid. If you do want to allow duplicates, then insertions are not a problem. You can just stick it either in the left subtree or the right subtree. The problem is that you can't count on the duplicates being on

问题 I'm writing a program that determines how many comparisons it takes to run a binary search algorithm for a given number and sorted array. What I don't understand is what counts as a comparison. // returns the number of comparisons it takes to find key in sorted list, array public static int binarySearch(int key, int[] array) { int left = 0; int mid; int right = array.length - 1; int i = 0; while (true) { if (left > right) { mid = -1; break; } else { mid = (left + right)/2; if (key < array[mid

问题 I'm writing a program that determines how many comparisons it takes to run a binary search algorithm for a given number and sorted array. What I don't understand is what counts as a comparison. // returns the number of comparisons it takes to find key in sorted list, array public static int binarySearch(int key, int[] array) { int left = 0; int mid; int right = array.length - 1; int i = 0; while (true) { if (left > right) { mid = -1; break; } else { mid = (left + right)/2; if (key < array[mid

问题 I have created a program which compares different search methods which search for a random int value 0-999 from a sorted array which is 0-999. I have created a binary search which works perfectly and after doing this I decided to try to create a search which, instead of splitting the values into half, splits them into 1/3 and 2/3 depending. So basically if I have {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15} and I was looking for 10 I would go from above to {6,7,8,9,10,11,12,13,14,15} to {10,11,12,13