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

Example 1:

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

Example 2:

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

Note:

1. 3 <= A.length <= 10000
2. 0 <= A[i] <= 10^6
3. A is a mountain, as defined above.

思路：就是這個array guarantee 是有mountain point的，題目只要找出來index。

利用性質來做binary search，A[mid-1] < A[mid] > A[mid+1], 這才是mountain point，如果落到左邊A[mid-1] < A[mid] , start = mid；如果落到右邊，A[mid] > A[mid+1] , end = mid;  

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