FindPeakElement.java

/*
 * 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.

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

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

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

Challenge
Time complexity O(logN)
 */
public class FindPeakElement {

    public int findPeak(int[] A) {
        int low = 0;
        int high = A.length - 1;
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (A[mid] < A[mid + 1]) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }

/*********************************************************/

    public int findPeak(int[] A) {
        for (int i = 1; i < A.length - 1; ++i) {
            if (A[i - 1] < A[i] && A[i] > A[i + 1]) {
                return i;
            }
        }
        return -1;
    }

}