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.

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

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

Time complexity O(logN)

halfhalf二分法（OOXXOOXX）。如果卡到爬坡知道右邊肯定存在peak，如果卡到降坡知道左邊肯定存在peak，所以每次還是可以果斷改start, end取半的。

public class Solution {
    /*
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    public int findPeak(int[] A) {
        // write your code here
        if (A == null || A.length == 0){
            throw new IllegalArgumentException();
        }

        int
 
 start = 0;
        int end = A.length - 1;

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

        if (A[start] > A[end]){
            return start;
        }
        return end;
    }
}

lintcode75- Find Peak Element- medium