class Solution: def getMaximumGenerated(self, n: int) -> int: if not n: return 0 if n == 1: return 1 nums = [0 for _ in range(n + 1)] nums[1] = 1 maxnum = 1 i = 1 while i < n // 2: nums[2 * i] = nums[i] nums[2 * i + 1] = nums[i] + nums[i + 1...
[LeetCode] 1646. Get Maximum in Generated Array You are given an integern. An arraynumsof lengthn + 1is generated in the following way: nums[0] = 0 nums[1] = 1 nums[2 * i] = nums[i]when2 <= 2 * i <= n nums[2 * i + 1] = nums[i] + nums[i + 1]when2 <= 2 *...
Can you solve this real interview question? Get Maximum in Generated Array - You are given an integer n. A 0-indexed integer array nums of length n + 1 is generated in the following way: * nums[0] = 0 * nums[1] = 1 * nums[2 * i] = nums[i] when 2 <= 2
Can you solve this real interview question? Maximum Value of a String in an Array - The value of an alphanumeric string can be defined as: * The numeric representation of the string in base 10, if it comprises of digits only. * The length of the strin
参考这个链接[LeetCode] Maximum XOR of Two Numbers in an Array 数组中异或值最大的两个数字 这个方法肯定想不到 代码如下: #include <iostream> #include <vector> #include <map> #include <unordered_map> #include <set> #include <unordered_set> ...
leetcode根本不让你通过的,比如100个数据100^100次循环 天知道要跑多久。 神奇的Kadane's algorithm 代码 class Solution: def maxSubArray(self, nums: List[int]) -> int: max_current = max_global = nums[0] for num in nums[1:]: max_current = max(num,max_current+num) if(max_current...
13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 参考文献 [LeetCode] Maximum XOR of Two Numbers in an Array 数组中异或值最大的两个数字【特别好】【位运算】maximum-xor-of-two-numbers-in-an-array
来自专栏 · LeetCode 1 人赞同了该文章 Description Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which has the largest product. Example 1: Input: [2,3,-2,4]Output: 6Explanation: [2,3] has the largest product 6.Example 2: In...
1. Description Get Maximum in Generated Array 2. Solution Version 1 classSolution:defgetMaximumGenerated(self,n:int)->int:ifn==0:return0ifn==1:return1maximum=1nums=[0for_inrange(n+1)]nums[1]=1forkinrange(2,n+1):i=k//2ifk%2==0:nums[k]=nums[i]else:nums[k]=nums[i]+nums[...
for i in range(len(row_pooling_T)): res.append(solution.maxSlidingWindow(row_pooling_T[i], max_pooling_window[0])) print("\nAfter max-pooling:") print(np.array(solution.transpose(res)))测试结果下一篇题解 滑动窗口最大值 评论(3) 排序:最热 ...