852. Peak Index in a Mountain Array

Let's call an array A a mountain if the following properties hold:

• A.length >= 3
• There exists some 0 < i < A.length - 1 such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]

Given an array that is definitely a mountain, return any i such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1].

Example 1:

Input: [0,1,0]
Output: 1

Example 2:

Input: [0,2,1,0]
Output: 1

//linear approach: Time: O(n), Space: O(1)
    public int peakIndexInMountainArray(int[] A) {
        if (A == null || A.length == 0) {
            return -1;
        }
        
        for (int i = 0; i < A.length - 1; i++) {
            if (A[i] > A[i + 1]) {
                return i;
            }
        }
        
        return -1;
    }

//binary search approach: Time: O(logn), Space: O(1)
    public int peakIndexInMountainArray(int[] A) {
        if (A == null || A.length == 0) {
            return -1;
        }
        
        int start = 0;
        int end = A.length - 1;
        
        while (start < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] < A[mid + 1]) {
                start = mid + 1;//注意start是指到mid + 1， 而end是指到mid
            } else if (A[mid] > A[mid + 1]) {
                end = mid;
            } else {
                return mid;//这里题目表示没有两个数相等的情况，但为了严谨可以加上这句
            }
        }
        
        return start;//返回start和end都可以，因为while停止的条件就是start = end了
    }
//注： test cases： 
//[0,1,0] 奇数情况
//[0,1,2,0] 偶数情况

 similar: 162. Find Peak Element