LintCode： Find a Peak

here is an integer array which has the following features:

    * The numbers in adjacent positions are different.

    * A[0] < A[1] && A[A.length - 2] > A[A.length - 1].

We define a position P is a peek if A[P] > A[P-1] && A[P] > A[P+1].

Find a peak in this array. Return the index of the peak.

Note

The array may contains multiple peeks, find any of them.

Example

[1, 2, 1, 3, 4, 5, 7, 6]

return index 1 (which is number 2)  or 6 (which is number 7)

 
class Solution {

public:

    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */

    int findPeak(vector<int> A) {

        int n = A.size();       
        int l = 0, r = n-1;

        while(l <= r)  {
            int m = (l+r)/2;
            if (A[m]> A[m-1] && A[m] >A[m+1]) return m;
            if (A[m] < A[m-1]) r = m-1;
            else if (A[m] < A[m+1]) l = m+1;
        }        

        return l;
    }