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

Note:

1. 3 <= A.length <= 10000
2. 0 <= A[i] <= 10^6
3. A is a mountain, as defined above.
   因為山峰的數字肯定是大於左邊的數字跟右邊的數字，所以用二分查找左右逼近找速度比較快。
   
   

    1 int peakIndexInMountainArray(int* A, int ASize) {
    2     int low=0,high=ASize,mid;
    
    3     while(low<high){
    4         mid=low+(high-low)/2;
    5         if(A[mid]>A[mid+1]) high=mid;
    6         else    low=mid+1;
    7         
    8     }
    9     return low;
   10 }

852. Peak Index in a Mountain Array