Peak Index in a Mountain Array

Easy

Let's call an arrayAamountain if the following properties hold:

• A.length >= 3

• There exists some 0 < i < A.length - 1such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]

Given an array that is definitely a mountain, return any i such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1].

Example 1:

Input: 
[0,1,0]
Output: 
1

Example 2:

Input: 
[0,2,1,0]
Output: 
1

Note:

1. 3 <= A.length <= 10000

2. 0 <= A[i] <= 10^6

3. A is a mountain, as defined above.

Solution

Linear Scan

class Solution {
    public int peakIndexInMountainArray(int[] A) {
        for (int i = 0; i < A.length; i++) {
            if (i > 0 && i < A.length - 1 && A[i] > A[i - 1] && A[i] > A[i + 1]) {
                return i;
            }
        }
        return -1;
    }
}

Linear Scan II

class Solution {
    public int peakIndexInMountainArray(int[] A) {
        int i = 0;
        while (A[i] < A[i+1]) i++;
        return i;
    }
}

Binary Search

BS Template #2

class Solution {
    public int peakIndexInMountainArray(int[] A) {
        int left = 0, right = A.length;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (A[mid] < A[mid + 1]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }
}

BS Template #1

class Solution {
    public int peakIndexInMountainArray(int[] A) {
        int left = 0, right = A.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (A[mid] < A[mid + 1]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }
}