162. Find Peak Element

package LeetCode_162

/**
 * 162. Find Peak Element
 * https://leetcode.com/problems/find-peak-element/description/
 *
 * A peak element is an element that is greater than its neighbors.
Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.
The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
You may imagine that nums[-1] = nums[n] = -∞.

Example 1:
Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.

Example 2:
Input: nums = [1,2,1,3,5,6,4]
Output: 1 or 5
Explanation: Your function can return either index number 1 where the peak element is 2,
or index number 5 where the peak element is 6.

Follow up: Your solution should be in logarithmic complexity.
 * */
class Solution {
    /*
    * solution 1: new array, Time complexity:O(n), Space complexity:O(n)
    * solution 2:binary search, Time complexity:O(logn), Space complexity:O(1)
    * */
    fun findPeakElement(nums: IntArray): Int {
        //solution 1
        //because may imagine that nums[-1] = nums[n] = -∞.
        val newArray = IntArray(nums.size + 2)
        //destPost=1, represent the position that put nums in newArray
        System.arraycopy(nums, 0, newArray, 1, nums.size)
        newArray.set(0,Int.MIN_VALUE)
        newArray.set(newArray.size-1,Int.MIN_VALUE)
        for (i in 1 until newArray.size - 1) {
            if (newArray[i] > newArray[i - 1] && newArray[i] > newArray[i + 1]) {
                return i
            }
        }

        //solution 2:
        var left = 0
        var right = nums.size - 1
        while (left <= right) {
            val mid = left + (right - left) / 2
            if ((mid - 1 < 0 || nums[mid] > nums[mid - 1]) && (mid + 1 >= nums.size || nums[mid] > nums[mid + 1])) {
                return mid
            } else if (nums[mid] < nums[mid + 1]) {
                //find in right side
                left = mid + 1
            } else {
                right = mid - 1
            }
        }
        return 0
    }
}