[LeetCode] 162. Find Peak Element

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Problem

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:

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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:

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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.

Note:

Your solution should be in logarithmic complexity.

Explanation

  1. Runtime is $\log(n)$, so we need to use binary search method to solve it, so we will create left and right pointers.

  2. When we find the middle element, we need to compare it with nums[mid+1]. If nums[mid] < nums[mid+1], then the peek element is on the right side because if nums[mid+1] > nums[mid+2], then the peek element is nums[mid+1], if nums[mid+1] < nums[mid+2], then we keep find the peek on the right side, which is nums[mid+2:end]. When left and right pointer meet, we return left or right.

Solution

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