标签：integer type multi func app str 最小 数组 ref
Description: A peak element is an element that is strictly greater than its neighbors.
Given an integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.
You may imagine that nums[-1] = nums[n] = -∞.
Link: 162. Find Peak Element
Examples:
Example 1: Input: nums = [1,2,3,1] Output: 2 Explanation: 3 is a peak element and your function should return the index number 2. Example 2: Input: nums = [1,2,1,3,5,6,4] Output: 5 Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.
思路: 顺序检索，从0，到最后一个元素，从第一个元素开始想，如果下一个小，则这个值就是局部峰值，返回，否则下一个有可能是，continue.
class Solution(object): def findPeakElement(self, nums): """ :type nums: List[int] :rtype: int """ n = len(nums) nums.append(float(‘-inf‘)) for i in range(n): if nums[i] > nums[i+1]: return i else: continue
但是上面的复杂度是O(n)，题目要求logn.怎么二分查找呢？二分查找的核心思想就是不断缩小带查找元素所在的区间，区间缩到最小返回。因为按题目意思，必然存在一个局部峰值，数组头和尾的两边设想存在最小负数，所以对于头，只要下一个元素小于它，它就是局部峰值，只要尾的前一个元素小于它，它就是局部峰值。当我们到了mid, 如果mid+1大于它，那么向右呈现上升趋势，峰值必然在mid右侧，l=mid+1，反之，呈现下降趋势，峰值必然在mid或者mid的左边,所以r=mid。以left < right结束，最后区间left=right，缩小到一个值，一个index,返回即可。
class Solution(object): def findPeakElement(self, nums): """ :type nums: List[int] :rtype: int """ n = len(nums) nums.append(float(‘-inf‘)) l, r = 0, n-1 while l < r: mid = int((l+r)/2) if nums[mid] > nums[mid+1]: r = mid else: l = mid+1 return l
日期: 2021-04-12 是阴天啊
Leetcode** 162. Find Peak Element
标签：integer type multi func app str 最小 数组 ref
原文地址：https://www.cnblogs.com/wangyuxia/p/14646447.html
