binary-search

Java equivalent of c++ equal_range (or lower_bound & upper_bound)

我只是一个虾纸丫 提交于 2019-11-27 06:04:21
问题 I have a List of object sorted and I want to find the first occurrence and the last occurrence of an object. In C++, I can easily use std::equal_range (or just one lower_bound and one upper_bound). For example: bool mygreater (int i,int j) { return (i>j); } int main () { int myints[] = {10,20,30,30,20,10,10,20}; std::vector<int> v(myints,myints+8); // 10 20 30 30 20 10 10 20 std::pair<std::vector<int>::iterator,std::vector<int>::iterator> bounds; // using default comparison: std::sort (v

Using Binary Search with sorted Array with duplicates [duplicate]

本小妞迷上赌 提交于 2019-11-27 05:26:54
问题 This question already has an answer here: Finding multiple entries with binary search 12 answers I've been tasked with creating a method that will print all the indices where value x is found in a sorted array. I understand that if we just scanned through the array from 0 to N (length of array) it would have a running time of O(n) worst case. Since the array that will be passed into the method will be sorted, I'm assuming that I can take advantage of using a Binary Search since this will be O

Searching for an element in a circular sorted array

拥有回忆 提交于 2019-11-27 05:13:04
问题 We want to search for a given element in a circular sorted array in complexity not greater than O(log n) . Example: Search for 13 in {5,9,13,1,3} . My idea was to convert the circular array into a regular sorted array then do a binary search on the resulting array, but my problem was the algorithm I came up was stupid that it takes O(n) in the worst case: for(i = 1; i < a.length; i++){ if (a[i] < a[i-1]){ minIndex = i; break; } } then the corresponding index of ith element will be determined

How to perform a binary search on IList<T>?

妖精的绣舞 提交于 2019-11-27 04:11:28
Simple question - given an IList<T> how do you perform a binary search without writing the method yourself and without copying the data to a type with build-in binary search support. My current status is the following. List<T>.BinarySearch() is not a member of IList<T> There is no equivalent of the ArrayList.Adapter() method for List<T> IList<T> does not inherit from IList , hence using ArrayList.Adapter() is not possible I tend to believe that is not possible with build-in methods, but I cannot believe that such a basic method is missing from the BCL/FCL. If it is not possible, who can give

Implement binary search in objects

左心房为你撑大大i 提交于 2019-11-27 03:35:53
Is there any way to implement binary search in a ArrayList with objects? In this example the ArrayList will be sorted with the field 'id'. class User{ public int id; public string name; } ArrayList<User> users = new ArrayList<User>(); sortById(users); int id = 66 User searchuser = getUserById(users,id); How would the "User getUserById( ArrayList users, int userid )" look like if I it should return the user with a specified id using binary search? Is this even possible? The Object Ordering article of The Java Tutorials has an example of writing your own Comparator in order to perform

Find a missing 32bit integer among a unsorted array containing at most 4 billion ints

一笑奈何 提交于 2019-11-27 02:55:49
问题 This is the problem described in Programming pearls . I can not understand binary search method descrbied by the author. Can any one helps to elaborate? Thanks. EDIT: I can understand binary search in general. I just can not understand how to apply binary search in this special case. How to decide the missing number is in or not in some range so that we can choose another. English is not my native language, that is one reason I can not understand the author well. So, use plain english please:

How is it possible to do binary search on a singly-linked list in O(n) time?

左心房为你撑大大i 提交于 2019-11-27 01:51:40
问题 This earlier question talks about doing binary search over a doubly-linked list in O(n) time. The algorithm in that answer work as follows: Go to the middle of the list to do the first comparison. If it's equal to the element we're looking for, we're done. If it's bigger than the element we're looking for, walk backwards halfway to the start and repeat. If it's smaller than the element we're looking for, walk forwards halfway to the start and repeat. This works perfectly well for a doubly

How can I better understand the one-comparison-per-iteration binary search?

回眸只為那壹抹淺笑 提交于 2019-11-27 01:25:47
问题 What is the point of the one-comparison-per-iteration binary search? And can you explain how it works? 回答1: There are two reasons to binary search with one comparison per iteration. The less important is performance. Detecting an exact match early using two comparisons per iteration saves an average one iteration of the loop, whereas (assuming comparisons involve significant work) binary searching with one comparison per iteration almost halves the work done per iteration. Binary searching an

how to calculate binary search complexity

回眸只為那壹抹淺笑 提交于 2019-11-26 23:26:19
I heard somebody say that since binary search halves the input required to search hence it is log(n) algorithm. Since I am not from a mathematics background I am not able to relate to it. Can somebody explain it in a little more detail? does it have to do something with the logarithmic series? duedl0r Here a more mathematical way of seeing it, though not really complicated. IMO much clearer as informal ones: The question is, how many times can you divide N by 2 until you have 1? This is essentially saying, do a binary search (half the elements) until you found it. In a formula this would be

The possible number of binary search trees that can be created with N keys is given by the Nth catalan number. Why?

自闭症网瘾萝莉.ら 提交于 2019-11-26 22:21:18
问题 This has been bothering me for a while. I know that given N keys to arrange in the form of a binary search tree, the possible number of trees that can be created correspond to the Nth number from the Catalan sequence. I have been trying to determine why this is; unable to find anything that might even attempt to explain it intuitively I resort to the collective knowledge of SO. I found other ways to calculate the number of possible trees, but they seemed less intuitive and no explanation was