Find Peak Element(lintcode 75)

Description

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)

Interface

class Solution {
    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    public int findPeak(int[] A) {
        // write your code here
    }
}

Solution

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) {
            return -1;
        }

        if (A.length == 1) {
            return 0;
        }

        int start = 1;
        int end = A.length - 2;

        while (start + 1 < end) {
            int mid = start + (end - start) / 2;

            if (A[mid] < A[mid - 1]) {
                end = mid;
            } else if (A[mid] < A[mid + 1]) {
                start = mid;
            } else {
                return mid;
            }
        }

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