Let's call an array arr a mountain if the following properties hold:
arr.length >= 3- There exists some
iwith0 < i < arr.length - 1such that:arr[0] < arr[1] < ... arr[i-1] < arr[i]arr[i] > arr[i+1] > ... > arr[arr.length - 1]
Given an integer array arr that is guaranteed to be a mountain, return any i such that arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].
Example 1:
Input: arr = [0,1,0] Output: 1
Example 2:
Input: arr = [0,2,1,0] Output: 1
Example 3:
Input: arr = [0,10,5,2] Output: 1
Example 4:
Input: arr = [3,4,5,1] Output: 2
Example 5:
Input: arr = [24,69,100,99,79,78,67,36,26,19] Output: 2
Constraints:
3 <= arr.length <= 1040 <= arr[i] <= 106arris guaranteed to be a mountain array.
Follow up: Finding the O(n) is straightforward, could you find an O(log(n)) solution?
山脉数组的峰顶索引。
符合下列属性的数组 arr 称为 山脉数组 ：
arr.length >= 3
存在 i（0 < i < arr.length - 1）使得：
arr[0] < arr[1] < ... arr[i-1] < arr[i]
arr[i] > arr[i+1] > ... > arr[arr.length - 1]
给你由整数组成的山脉数组 arr ，返回任何满足 arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1] 的下标 i 。来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/peak-index-in-a-mountain-array
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影子题162。这个题还有一个followup，1095题。题目说的很绕，其实是给了一个数组，其中只有一个峰值，峰值左半边递增，右半边递减，请你找出这个峰值。
两种思路，一种是线性找，找if (nums[i] > nums[i - 1])，一种是二分法。我这里给出二分法的代码。
时间O(logn)
空间O(1)
Java实现
1 class Solution { 2 public int peakIndexInMountainArray(int[] arr) { 3 int left = 0; 4 int right = arr.length - 1; 5 while (left < right) { 6 int mid = left + (right - left) / 2; 7 if (arr[mid] < arr[mid + 1]) { 8 left = mid + 1; 9 } else { 10 right = mid; 11 } 12 } 13 return left; 14 } 15 }
JavaScript实现
1 var peakIndexInMountainArray = function (arr) { 2 let left = 0; 3 let right = arr.length; 4 while (left < right) { 5 let mid = Math.floor(left + (right - left) / 2); 6 if (arr[mid] < arr[mid + 1]) { 7 left = mid + 1; 8 } else { 9 right = mid; 10 } 11 } 12 return left; 13 };