[Binary Search] 162. Find Peak Element

문제

A peak element is an element that is strictly greater than its neighbors.
Given a 0-indexed integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.
You may imagine that nums[-1] = nums[n] = -∞. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.
You must write an algorithm that runs in O(log n) time.
 
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: 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.
 
Constraints:
1 <= nums.length <= 1000
-2^31 <= nums[i] <= 2^31 - 1
nums[i] != nums[i + 1] for all valid i.

 

문제접근

O(log n) 의 시간 복잡도를 가질려면 완전 탐색이 아닌 다른 접근법을 생각한다.

1. peak는 양 옆 보다 큰 수를 찾으면된다. (가장 큰수를 찾으면 해결)

2. 이분탐색을 통해 가장 큰 수를 반환한다.

 

문제 풀이

class Solution {
    public int findPeakElement(int[] nums) {
        int start = 0;
        int end = nums.length - 1;

        while (end > start) { 
            int mid = (start + end) / 2;
            if (nums[mid] < nums[mid + 1]) 
                start = mid + 1;
             else
                 end = mid;

        }
        return end;
    }
}