difficult：easy #852
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.

解法一:

#include<vector>
int peakIndexInMountainArray(std::vector<int>& A) {
    int start = (A.size() - 1) / 2;
    while (start != 0 || start != A.size() - 1) {
        if (A[start - 1] < A[start] && A[start] < A[start + 1])
            start++;
        else if (A[start - 1] > A[start] && A[start] > A[start + 1])
            start--;
        else if (A[start - 1] <= A[start] && A[start] >= A[start + 1])
            return start;
    }
}

解法二(二分查找):

int peakIndexInMountainArray(std::vector<int>& A) {
    int start = 0, end = A.size() - 1;
    while (start < end){
        int mid = start +(end-start) / 2;
        if (A[mid] < A[mid + 1])
            start = mid + 1;
        else
            end = mid;
    }
    return start;
}

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