Lintcode: Find a Peak

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

 1 class Solution {

 2     /**

 3      * @param A: An integers array.

 4      * @return: return any of peek positions.

 5      */

 6     public int findPeak(int[] A) {

 7         if (A==null || A.length<3) return -1;

 8         int l = 1;

 9         int r = A.length - 2;

10         while (l <= r) {

11             int m = (l + r) / 2;

12             if (A[m]>A[m+1] && A[m]>A[m-1]) return m;

13             else if (A[m]<A[m+1] && A[m]>A[m-1]) {

14                 l = m + 1;

15             }

16             else {

17                 r = m - 1;

18             }

19         }

20         return -2;

21     }

22 }