Find Peak Element
A peak element is an element that is strictly greater than its neighbors. Given a 0-indexed 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] = -∞`. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array. You must write an algorithm that runs in `O(log n)` time.
Examples
Constraints
1 <= nums.length <= 1000-2^31 <= nums[i] <= 2^31 - 1nums[i] != nums[i + 1] for all valid i.
Approach
Iterate through the array and check if the current element is greater than its neighbors. Since `nums[-1]` and `nums[n]` are considered negative infinity, we only need to check if the current element is greater than the next element. The first such element we find is a peak.
Complexity Analysis
This approach does not meet the O(log n) time complexity requirement.
class Solution { public int findPeakElement(int[] nums) { for (int i = 0; i < nums.length - 1; i++) { if (nums[i] > nums[i + 1]) { return i; } } return nums.length - 1; }}