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

• It's guaranteed the array has at least one peak.
• The array may contain multiple peeks, find any of them.
• The array has at least 3 numbers in it.

Have you met this question in a real interview?

Yes

Example

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

Return index1(which is number 2) or6(which is number 7)

Challenge

Time complexity O(logN)

LeetCode: https://leetcode.com/problems/find-peak-element/description/\#

       public int FindPeakElement(int[] nums) {

            int start = 0, end = nums.Length - 1;
            while (start + 1 < end) {
                int mid = start + (end - start) / 2;
                if (nums[mid] > nums[mid - 1] && nums[mid] > nums[mid + 1]){
                    start = mid;
                    break;
                }
                else if (nums[mid] > nums[mid - 1])
                {
                    start = mid;

                }
                else if (nums[mid] < nums[mid + 1])
                {
                    end = mid;
                }

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

Answer2 (from Jiuzhang)

    public int findPeak(int[] A) {
        // write your code here
        int start = 1, end = A.length-2; // 1.答案在之间，2.不会出界 
        while(start + 1 <  end) {
            int mid = (start + end) / 2;
            if(A[mid] < A[mid - 1]) {
                end = mid;
            } else if(A[mid] < A[mid + 1]) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if(A[start] < A[end]) {
            return end;
        } else { 
            return start;
        }
    }