binary-search

问题 I want to modify the famous binary search algorithm to return the index of the next bigger item instead of the key being searched. So we have 4 cases: the key is smaller than all items, return 0. the key is bigger than all items, return items.length. the key is found at index x, return x+1. the key isn't found, return the index of the next bigger one. e.g: data = { 1, 3, 5, 7, 9, 11 }; search for 0 returns 0. search for 11 or 12 returns 6. search for 5 or 6 returns 3. while (low <= high) {

问题 I'm looking at List and I see a BinarySearch method with a few overloads, and I can't help wondering if it makes sense at all to have a method like that in List? Why would I want to do a binary search unless the list was sorted? And if the list wasn't sorted, calling the method would just be a waste of CPU time. What's the point of having that method on List? 回答1: Sorting and searching are two very common operations on lists. It would be unfriendly to limit a developer's options by not

问题 Two sorted arrays of integers A[1..N] and B[1..N] are provided in ascending order . Q:Design an O(log N)-time algorithm for finding out the median of all 2N integers. N may not power of 2 . To make thing easy, we can assume O(1) algorithm which return m such that: 2^m < N < 2^m+1 My problems: N may not be power of 2 , what does that mean? (understood) I don't know how to change the input and make the length to power of 2 after found min and max elements from both array A and B . 回答1: You can

问题 I came across this problem - input - I am given two sorted arrays a1 and a2. I need to find the elements that are not present in the second array. I have two approaches 1) Hashtable - O(m+n) [use when second array is small] 2) Binary Search - O(m*logn) [use when second array is huge] Are there any other approaches with better time complexities? Thank You 回答1: Just iterate them in parallel. Here's a JavaScript example: var a1 = [1, 2, 3, 4, 5, 6, 7, 9]; var a2 = [0, 2, 4, 5, 8]; findNotPresent

问题 I have used this algorithm many times to binary search over Ints or Longs . Basically, I start from Long.MinValue and Long.MaxValue and decide to set the bit at i th position depending on the value of the function I am maximizing (or minimizing). In practice, this turns out to be faster (exactly 63*2 bitwise operations) and easier to code and avoids the many gotchas of traditional binary search implementations. Here is my algorithm in Scala: /** * @return Some(x) such that x is the largest

问题 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 9 years ago . how to find a random element in a sorted array of unknown length. 回答1: I'll assume you mean how do I find if an element is part of the array? not how do I return a random element from the array? . Use binary

问题 I wanted to find out the trade off point between a Dictionary lookup and a binary search lookup of an array. I was expecting constant time lookups for the Dictionary, and logarithmic time lookups for the binary search depending on the size of the collection, with the binary search performing better for smaller sized collections. However, I was surprised when I saw the following results: I was surprised at: 1. Binary search is growing logarithmically at first, and then grows much faster. 2.

问题 I am trying to tackle a classical interview problem which is basically performing a binary search on a list which first increases then decreases. Even though it's obvious that we can achieve O(log n) I couldn't figure out what is wrong the code below that I've written: #include <iostream> using namespace std; int binarySearch(int *A, int low, int high, int key) { while(low < high) { int mid = (low + high) / 2; if(key < A[mid]) { if(A[mid - 1] < A[mid] && A[mid] < A[mid + 1]) high = mid - 1;

问题 I always have the hardest time with this and I have yet to see a definitive explanation for something that is supposedly so common and highly-used. We already know the standard binary search. Given starting lower and upper bounds, find the middle point at (lower + higher)/2, and then compare it against your array, and then re-set the bounds accordingly, etc. However what are the needed differences to adjust the search to find (for a list in ascending order): Smallest value >= target Smallest

问题 I have a binary search loop which gets hit many times in the execution path. A profiler shows that the division part of the search (finding the middle index given the high and low indices of the search range) is actually the most costly part of the search, by a factor of about 4. (I think) it is not critical for efficient binary search to find the exact middle value, just a value near the middle which does not have bias in either direction. Is there a bit-twiddling algorithm to replace mid =