leetcode162. Looking for peak

Problem description ：

The peak element refers to the element whose value is strictly greater than the left and right adjacent values .

Give you an array of integers nums, Find the peak element and return its index . The array may contain multiple peaks , under these circumstances , return Any peak Just where you are .

You can assume nums[-1] = nums[n] = -∞ .

You must achieve a time complexity of O(log n) Algorithm to solve this problem .

Example ：

Example 1：

Input ：nums = [1,2,3,1]
Output ：2
explain ：3 Is the peak element , Your function should return its index 2.

Example 2：

Input ：nums = [1,2,1,3,5,6,4]
Output ：1 or 5
explain ： Your function can return the index 1, Its peak element is 2;
     Or return index 5, Its peak element is 6.

Tips ：

    1 <= nums.length <= 1000
    -231 <= nums[i] <= 231 - 1
    For all that works i There are nums[i] != nums[i + 1]

Code up , Take it to run ：

package com.onlyqi.test03.erfen;

/**
 * @author onlyqi
 * @date 2022/7/12 10:21
 * @description
 */
public class Test04 {
    public static void main(String[] args) {
        int[]nums = {1,2,1,3,5,6,4};
        System.out.println(findPeakElement(nums));
    }
    public static int findPeakElement(int[] nums) {
        int start = 0, end = nums.length - 1;
        while (start < end ) {
            int mid = (start + end ) / 2;
            if (nums[mid] > nums[mid + 1])
                end = mid;
            else
            start = mid + 1;
        }
        return start;
    }

    /**
     *  Easy to think of Solutions 
     * @param nums
     * @return
     */
    public static int findPeakElement1(int[] nums) {
        for (int i = 1; i < nums.length; i++) {
            if(nums[i]>nums[i-1]&&nums[i]>nums[i+1]){
                return i;
            }
        }
        return 0;
    }

}

Running results ：

I want to brush it 300 Algorithm questions , The first 105 Avenue