75. find peak element

public class  {
    
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    public int findPeak(int[] A) {
        // write your code here

        
        保证array中有peak
        The numbers in adjacent positions are different.
        A[0] < A[1] && A[A.length - 2] > A[A.length - 1].


        找到任意一个peak, return index
        */

        int start = 0, end = A.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] < A[mid - 1]) {
                end = mid; // 不能排除mid
            } else {
                start = mid;
            }
        }
        if (A[start] < A[end]) {
            return end;
        } else {
            return start;
        }

    }
}