http://www.lintcode.com/en/problem/find-peak-element/

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.

Notice It's guaranteed the array has at least one peak. The array may contain multiple peeks, find any of them. The array has at least 3 numbers in it.

Example Given [1, 2, 1, 3, 4, 5, 7, 6] Return index 1 (which is number 2) or 6 (which is number 7)

解题：从数组里面找个peak，二分法检查部分也就是检查看左右两边的值是否比当前值小。不然就看情况更新end或start

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


    public int findPeak(int[] A) {
        if (A == null && A.length == 0) {
            return -1;
        }

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