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
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[...
这时候其实往左往右跳都可以。随便指定一个方向都可以,要么往左找到2,要么往右找到7(根据定义它也是peak value)。 因为只用找出一个,所以还是O(logN) 1publicclassSolution {2publicintfindPeakElement(int[] num) {3intl = 0;4intr = num.length - 1;5while(l <=r) {6if(l == r)returnl;7intmid...
AI代码解释 classSolution{publicintfindPeakElement(int[]nums){int idx=0;for(int i=1;i<nums.length;++i){if(nums[i]>nums[idx]){idx=i;}}returnidx;}} 3、时间复杂度 时间复杂度:O(n) 其中n是数组nums的长度。 空间复杂度:O(1) 只需要常量级的指数空间。 三、总结 只要数组中存在一个元素比...
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 ...
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. ...
A peak element is an element that is greater than its neighbors. Given an input array where 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. ...
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...
const findPeakElement = nums => { let [left, right] = [0, nums.length - 1]; while (left < right) { const mid = left + (right - left) >> 1;//不断二分 寻找上升元素对 if (nums[mid] > nums[mid + 1]) { right = mid;//下降 } else { left = mid + 1;//上升 } } ...