A peak element is an element that is greater than its neighbors.
Given an input array where num[i] ≠ num[i+1], find a peak element and return its index.
The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
You may imagine that num[-1] = num[n] = -∞.
For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.

Solution1：Binary Search (Iterative)

思路： if nums[i] > nums[i + 1]瘫想，為" \ " Pattern，因?yàn)樽笥覂蛇厼樨?fù)無窮，所以此pattern左邊肯定含有peak, 則向二分左邊部分繼續(xù)找。else when if nums[i] < nums[i + 1]，為" / " Pattern初烘，所以此pattern右邊肯定含有peak, 則向二分右邊部分繼續(xù)找义桂。
Time Complexity: O(logN) Space Complexity: O(1)

Solution2：Binary Search (Recursive)

思路： 思路同1的遞歸寫法
Time Complexity: O(logN) Space Complexity: O(1)

Solution3：Binary Search Template

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

Solution1 Code：

class Solution1 {
    public int findPeakElement(int[] nums) {
        int start = 0, end = nums.length - 1;
        
        while(start < end) {
            int mid1 = (start + end) / 2;
            int mid2 = mid1 + 1;
            
            if(nums[mid1] < nums[mid2]) {
                // found "/" pattern, so the peak must be on the right
                start = mid2;
            }
            else {
                // found "\" pattern, so the peak must be on the left
                end = mid1;
            }
        }
        return start;
    }
}

Solution2 Code：

class Solution2 {
    public int findPeakElement(int[] nums) {
        return helper(nums, 0, nums.length - 1);
    }
    
    private int helper(int[] nums, int start, int end) {
        if(start == end) {
            return start;
        }
        else {
            int mid1 = (start + end) / 2;
            int mid2 = mid1 + 1;
            if(nums[mid1] < nums[mid2]) {
                // found "/" pattern, so the peak must be on the right
                return helper(nums, mid2, end);
            }
            else {
                // found "\" pattern, so the peak must be on the left
                return helper(nums, start, mid1);
            }
        }
    }
}

Solution3 Binary Search Template Code：

class Solution {
    public int findPeakElement(int[] nums) {
        if(nums == null || nums.length == 0) return 0;
        
        int left = 0, right = nums.length - 1;
        
        while(left + 1 < right) {
            int mid = left + (right - left) / 2;
            if(nums[mid] > nums[mid - 1]) { 
                // "/"
                left = mid;
            }
            else {
                right = mid;
            }
        }
        
        if(nums[left] > nums[right]) {
            return left;
        }
        else {
            return right;
        } 
    }
}