binary-search

问题 I've read multiple articles including Jon Bentleys chapter on binary search. This is what I understand about CORRECT binary search logic and it works in the simple tests I did: binarysearch (arr, low, high, k) 1. while (low < high) 2. mid = low + (high - low)/2 3. if (arr[mid] == k) return mid 4. if (arr[mid] < k ) high = mid -1 5. else low = mid + 1 Now to find the 1st occurence with sorted duplicates, you'd chance line 3 if condition to continue instead of returning mid as binarysearch_get

问题 When implementing Insertion Sort, a binary search could be used to locate the position within the first i - 1 elements of the array into which element i should be inserted. How would this affect the number of comparisons required? How would using such a binary search affect the asymptotic running time for Insertion Sort? I'm pretty sure this would decrease the number of comparisons, but I'm not exactly sure why. 回答1: Straight from Wikipedia: If the cost of comparisons exceeds the cost of

问题 When addressed the Binary_Search in chapter 2(2.4.4),the author mentioned that " Notice that the variables cannot be declared unsigned(why?).In cases where the unsigned qualifier is questionable, we will not use it. As an example, if the unsigned qualifier is dependent on an array not beginning at zero, we will discard it. we will not use it. As an example, if the unsigned qualifier is dependent on an array not beginning at zero, we will discard it. We will also avoid using the unsigned type

问题 When addressed the Binary_Search in chapter 2(2.4.4),the author mentioned that " Notice that the variables cannot be declared unsigned(why?).In cases where the unsigned qualifier is questionable, we will not use it. As an example, if the unsigned qualifier is dependent on an array not beginning at zero, we will discard it. we will not use it. As an example, if the unsigned qualifier is dependent on an array not beginning at zero, we will discard it. We will also avoid using the unsigned type

问题 So I am searching a huge ArrayList for a particular String value, but I need Collections.binarySearch() to return a value >=0 if the String I am looking for is equal( non case-sensitive) to the String I pass into the binarySearch() method. Now in the source code for Collections.binarySearch(), it eventually calls the following lines of code. Comparable<? super T> midVal = list.get(mid); int cmp = midVal.compareTo(key); So seen as I cannot override String as its final (therefore preventing me

问题 So I have to make a method which is using the quicksort as a sort algorithm without using the Java API. Then I have to write another method that gives the sorted array back and using the binary search return true if the searched element is found in it. I have no idea where am I making the stupid mistake. public class Aufgabe1 { public static void sort(int[] array) { /* TODO: add code here */ sort(array, 0, array.length - 1); } public static void sort(int[] array, int start, int end) { int i =

问题 I use an ArrayList for my binary search. The datagridview's rows is added to the ArryList. When I deleting a single row from the datagridview, it works almost perfectly. The problem is when I delete many rows from the datagridview from the top or the bottom and middle, it gives me an error. How can I refresh or update the ArrayList after I deleted a row from the ArrayList (datagridview)? The error: 'Index was out of range. Must be non-negative and less than the size of the collection.

问题 I wonder, can binary search be applied on a 2D array ? What would the conditions on the array be? Sorted on 2D?? What would be the time complexity for it? How would the algorithm change the boundary of the search (minX,maxX,minY,maxY) ?? Edit: Binary Search on 1D maintains 2 pointers minX and maxX .. It selects the middle index (minX+maxX)/2 and compare it with the search value, if greater then change maxX , else change minX ... until minX>=maxX Pseudo code for normal binary seacrh: min := 1;

问题 I wrote a function that returns the first met of each numbers from 0 to 9 array = [0,0,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9] def lower(a, val, left, right): if left == right: return left mid = (left + right) // 2 if a[mid] < val: return lower(a, val, mid+1, right) else: return lower(a, val, left, mid) for i in range(10): print(lower_bound(array, i , 0, len(array)-1), end =' ') result: 0 2 4 6 8 10 12 14 16 18 . I tried to write a function that return the last met of each numbers from 0 to 9.

问题 I wrote a function that returns the first met of each numbers from 0 to 9 array = [0,0,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9] def lower(a, val, left, right): if left == right: return left mid = (left + right) // 2 if a[mid] < val: return lower(a, val, mid+1, right) else: return lower(a, val, left, mid) for i in range(10): print(lower_bound(array, i , 0, len(array)-1), end =' ') result: 0 2 4 6 8 10 12 14 16 18 . I tried to write a function that return the last met of each numbers from 0 to 9.