A peak element is an element that is greater than its neighbors. num[i] ≠ num[i+1], find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. num[-1] = num[n] = -∞.[1, 2, 3, 1], ...
A peak element is an element that is greater than its neighbors.Given an input arraynums, wherenums[i] ≠ nums[i+1], find a peak element and return its index.The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. You may imagine that ...
https://leetcode.com/problems/find-peak-element/ 思路就是二分查找。如果mid在波峰的递减部分,那么e = mid - 1. 如果mid在波峰的递增部分,那么s = mid + 1. 如果mid在波峰,那么返回。 my code: 效率不高。 class Solution(object): def findPeakElement(self, nums): """ :type nums: List[int]...
A peak element is an element that is greater than its neighbors. Given an input array wherenum[i] ≠ num[i+1], find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. You may imagine thatnum[...
A peak element is an element that is greater than its neighbors.Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.The array may contain multiple peaks, in that case return the index to any one of the peaks is fine....
https://leetcode-cn.com/problems/find-peak-element/ 题目描述 峰值元素是指其值大于左右相邻值的元素。 给定一个输入数组nums,其中nums[i] ≠ nums[i+1],找到峰值元素并返回其索引。 数组可能包含多个峰值,在这种情况下,返回任何一个峰值所在位置即可。
这种找Peak Element的首先想到的是binary search, 因为有更优的时间复杂度,否则brute forceO(n)的方法太直接了。 binary search的方法就是比较当前element与邻居,至于左邻居还是右邻居都可以,只要一致就行。不断缩小范围,最后锁定peak element. 这种类似题也有很多Follow up, 比如先增后减的数组怎么找指定元素,甚至...
Can you solve this real interview question? Find Peak Element - A peak element is an element that is strictly greater than its neighbors. Given a 0-indexed integer array nums, find a peak element, and return its index. If the array contains multiple pea
C++ 智能模式 1 2 3 4 5 6 class Solution { public: int findPeakElement(vector<int>& nums) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2 nums = [1,2,3,1] 1 2 [1,2,3,1] [1,2,1,3,5,6,4] Source ...
Leetcode 218 The Skyline Problem Leetcode 480. Sliding Window Median (这个题用TreeMap超级方便) Leetcode 318 Count of Smaller Numbers After Self (这个题线段树、二分索引树、TreeMap都可以) 动态规划(Dynamic Programming) 基础知识:这里指的是用for循环方式的动态规划,非Memoization Search方式。DP可以在多...