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

Note:

3 <= A.length <= 10000
0 <= A[i] <= 10^6

A is a mountain, as defined above.
因为山峰的数字肯定是大于左边的数字跟右边的数字，所以用二分查找左右逼近找速度比较快。

 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 }