Find Peak Element

题目：

There 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 element in this array. Return the index of the peak.

分析：

二分让A[m-1] > A[m]和A[m] < A[m+1]时候移动，剩下情况A[m-1] <A[m] >A[m+1]就是峰值。

解法：

    public int findPeak(int[] A) {
        // write your code here
        if(A.length == 0)
        return -1;

        int s = 1, e = A.length - 2;//起点和终点位置只会在中间

        while(s + 1 < e){
            int m = s + (e - s)/2;

            if(A[m-1] > A[m]){
                e = m;
            }else if(A[m] < A[m+1]){
                s = m;
            }else{
                e = m; //正好是峰值，移动e好保留s
            }
        }

        if(A[s-1] < A[s] && A[s+1] < A[s])
            return s;

        if(A[e-1] < A[e] && A[e+1] < A[e])
            return e;

            return -1;
    }