Binary Search (Without Recursion): 337ms Binary Search (With Recursion): 514ms Process finished withexitcode 0 比较非递归和递归写法的二分查找的效率 非递归算法在性能上有微弱优势
如果待查找序列 seq 中有多个元素与 key 相等,那么,binary_search 函数只是返回其中一个元素的索引。比如,序列 seq 为 [1,2,5,5,5,5,7,8,9], key 的值为 5。 中间元素 seq[4] == 5,所以结果为 4。如果要返回从左到右第一个与 key 相等的元素的索引(此索引为 2),该如何做呢? 我们回想一下 ...
defbinary_search_recursion(lst, value, low, high):ifhigh <low:returnNone mid= (low + high) // 2iflst[mid] >value:returnbinary_search_recursion(lst, value, low, mid-1)eliflst[mid] <value:returnbinary_search_recursion(lst, value, mid+1, high)else:returnmid 2、循环的方法实现二分法 de...
# Python 3 program for recursive binary search.# Returns index of x in arr if present, else NonedefbinarySearch_recursion(arr:list,left,right,x):# base caseifleft<=right:mid=(left+right)//2# if element is smaller than mid, then it can only# be present in left subarrayifx<arr[mid]:...
Binary Trees With Factors- DP 3 -- 11:58 App leetcode-1081. Smallest Subsequence of Distinct Characters -dict-and-stack 79 -- 16:17 App leetcode-287-set-indexing-Floyd-trim 81 -- 11:30 App leetcode-2925. Maximum Score After Applying Operations on a Tree - recursion 81 -- 10:09...
Binary Search The last page covered some simple examples of recursion, which could have been coded with loops just as easily. In other cases, using recursion is considerably more 'natural' than using loops. The Algorithm You are given an array sorted in increasing order and would like to ...
The binary search begins by comparing the searched element with the middle element of the array. Since ‘mid’ calculate for every iteration or recursion, we divide the array into half and then try to solve the problem. If the searched value is less than the element in the middle of the ...
However, there are also certain risks involved with recursion, which is one of the subjects of the next section.Covering Tricky Details Here’s what the author of The Art of Computer Programming has to say about implementing the binary search algorithm: “Although the basic idea of binary ...
key: root.right = self.__insert(root.right, key) else: print key, 'is already in tree' return root ##non-recursion ## def insert(self, key): ## if not self.root: ## self.root = tree_node(key) ## else: ## cur = self.root ## while True: ## if key < cur.key: ## ...
data[1] = 3, and first is not greater than or equal to last. You thus don't hit either of the recursion terminating return calls. You value of 2 is less than data[1] of 3, so you call bsearch_recur again with the following parameters: 0, 1, 2. ...