Lintcode: 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 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)

Challenge
Time complexity O(logN)

跟Leetcode Find Peak Element一样

有一些考虑：因为梯度下降法是要比较m、m+1、m-1三个index大小，因此为保证不outofbound，令l = 1, r = A.length-2; 这样也可以maintain一个性质：l、r始终在peak element可能的区域内

class Solution {
/**
* @param A: An integers array.
* @return: return any of peek positions.
*/
public int findPeak(int[] A) {
if (A==null || A.length<3) return -1;
int l = 1;
int r = A.length - 2;
while (l <= r) {
int m = (l + r) / 2;
if (A[m]>A[m+1] && A[m]>A[m-1]) return m;
else if (A[m]<A[m+1] && A[m]>A[m-1]) {
l = m + 1;
}
else {
r = m - 1;
}
}
return -2;
}
}