LintCode Find Peak Element（查找峰值）

原题网址：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

The array may contains multiple peeks, find any of them.

Example

Given

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

Return index

1

(which
is number 2) or

6

(which is number 7)

方法：二分法。答案可能不唯一，只需要得到一个即可。

class Solution {
/**
* @param A: An integers array.
* @return: return any of peek positions.
*/
public int findPeak(int[] A) {
// write your code here
int i = 1, j = A.length - 2;
while (i <= j) {
int m = (i + j) / 2;
if (A[m - 1] < A[m] && A[m] > A[m + 1]) return m;
if (A[m - 1] < A[m]) i = m + 1; else j = m - 1;
}
return i;
}
}