classSolution {public:intfindPeakElement(vector<int>&nums) {for(inti =1; i < nums.size(); ++i) {if(nums[i] < nums[i -1])returni -1; }returnnums.size() -1; } }; Java 解法二: publicclassSolution {publicintfindPeakElement(int[] nums) {for(inti = 1; i < nums.length; ++i...
java代码如下: 1publicclassSolution {2publicintfindPeakElement(int[] num) {3intstart=0,end=num.length-1,mid=0,mid1=0;4while(start<end){5mid=(start+end)/2;6mid1=mid+1;7if(num[mid]<num[mid1]) start=mid1;8elseend=mid;9}10returnstart;11}12} 2、python: 1classSolution:2#@param...
Note: Your solution should be in logarithmic complexity. 原题链接 顺序遍历 复杂度 时间O(N) 空间 O(1) 思路 最简单的做法,遍历数组找出比前后都大的元素。 代码 public class Solution { public int findPeakElement(int[] nums) { for(int i = 1; i < nums.length-1; i++){ if(nums[i] > ...
Your solution should be in logarithmic complexity. 思路: 可能包含多个极值点(peak),只要取任意一个就行。假设只有一个peak,用二分查找根据mid元素单增单减的关系可以判断peak在mid的左边还是右边。如果不止一个peak,用二分法划分也不会将查找区间[left,right]划到没有peak的区间。时间复杂度为O(logn)。 算法:...
这个题 LintCode 和 LeetCode 的 find peak element 是有区别的。数据上,LintCode 保证数据第一个数比第二个数小,倒数第一个数比到倒数第二个数小。因此 start, end 的范围要取 1, A.length - 2 二分法。每次取中间元素,如果大于左右,则这就是peek。否则取大的一边,两个都大,可以随便取一边。最终会找...
Leetcode: Find Peak Element 题目: 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...
方法都是一样的,就是先得找到peak element的点,然后根据peak点将整个数组分成几部分,对于每个部分来说,是单调的,所以可以对每个部分分别用binary search来找元素。 复杂度 time: O(logn), space: O(1) 代码 public class Solution { public int findPeakElement(int[] nums) { ...
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
LeetCode 162. Find Peak Element 简介: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。 Description A peak element is an element that is greater than its neighbors....
:pencil2: 算法相关知识储备 LeetCode with Python :books:. Contribute to cancanxinxin/leetCode development by creating an account on GitHub.