BinarySearch[3]Leetcode852. Peak Index in a Mountain Array_there exists some index i with 0 < i

本文介绍如何高效地在给定的保证为山峰数组的整数数组中找到峰值索引，通过对比两种方法：暴力遍历和二分搜索。暴力遍历时间复杂度为O(n)，而二分搜索优化至O(log(n))。

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Description[Easy]

Let’s call an array arra mountain if the following properties hold:

arr.length >= 3
There exists some i with 0 < i < arr.length - 1 such 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].

Solution1: Brute force

Time Complexity: O(n)
Space Complexity: O(1)

class Solution {
    public int peakIndexInMountainArray(int[] arr) {
        //Brute force
        //Time Complexity: O(n)
        for(int i = 0; i < arr.length-1; ++i ){
            if( arr[i]> arr[i+1]){
                return i;
            }
        }
        return -1;
    }
}

Solution2: Binary Search

Time Complexity: O(log(n))
Space Complexity: O(1)

class Solution {
    public int peakIndexInMountainArray(int[] arr) {
        int start = 0, end = arr.length -1;
        
        while(start + 1 < end){
            int mid = start + (end - start)/2;
            if(arr[mid-1] < arr[mid]){
                start = mid;
            }
            else{
                end = mid;
            }
        }
        
        if(arr[start] < arr[end]) return end;
        return start;
    }
}