2、问题分析 直接从后向前遍历,找到 A[i] > A[i-1] 即可。 3.代码 1intpeakIndexInMountainArray(vector<int>&A) {2inti = A.size() -1;3while( i >0){4if( A[i] > A[i-1]){5returni;6}7i--;8}9}
Find Peak Element 参考资料: https://leetcode.com/problems/peak-index-in-a-mountain-array/ https://leetcode.com/problems/peak-index-in-a-mountain-array/discuss/139848/C%2B%2BJavaPython-Better-than-Binary-Search LeetCode All in One 题目讲解汇总(持续更新中...)...
思路 A[i]为数组A的最大值,然后返回其在A中的索引即可 代码实现 class Solution: def peakIndexInMountainArray(self, A): """ :type A: List[int] :rtype: int """ return A.index(max(A)) 1. 2. 3. 4. 5. 6. 7.
class Solution(object): def peakIndexInMountainArray(self, arr): """ :type arr: List[int] :rtype: int """ n = len(arr) left,right = 1,n-2 while left <= right: mid = left + (right - left)//2 if arr[mid] > arr[mid-1] and arr[mid] > arr[mid+1]: return mid elif...
def peakIndexInMountainArray(arr: list[int]) -> int: for i in range(1, len(arr)): # 注意:这里要从1 开始,否则再i-1的时候会有bug if arr[i] < arr[i - 1]: return i - 1 二分查找 当然,我们这是一道二分查找题,使用二分查找可以将时间复杂度降到最低 :O(log n) 回忆下二分查找...
【leetcode】山脉数组的峰顶索引,intpeakIndexInMountainArray(int*arr,intarrSize){inti=0;for(i=1;arr[i]>arr[i-1];i++);returni-1;}
C++实现LeetCode（162.求数组的局部峰值）[LeetCode] 162.Find Peak Element 求数组的局部峰值 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 peak...
给定一个长度为n的整数山脉数组arr,其中的值递增到一个峰值元素然后递减。 返回峰值元素的下标。 你必须设计并实现时间复杂度为O(log(n))的解决方案。 示例1: 输入:arr = [0,1,0]输出:1 示例2: 输入:arr = [0,2,1,0]输出:1 示例3: 输入:arr = [0,10,5,2]输出:1 ...
leetcode-34. Find First and Last Position of Element in Sorte-binary-searchyjhycl 立即播放 打开App,流畅又高清100+个相关视频 更多 84 0 18:47 App leetcode-222-Counter Complete Binary Tree Nodes - binary search 2 0 10:00 App leetcode-852. Peak Index in a Mountain Array -binary-search...
852Peak Index in a Mountain ArrayPythonJava1. Scan the array until encountering decline, O(n) and O(1) 2. Binary seach with additional check for [i + 1], O(logn) and O(1) 867Transpose MatrixPythonJavaRes[i][j] = A[j][i] ...