deeprun

Find Peak Element

 문제 설명

A peak element is an element that is strictly greater than its neighbors.

Given an 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] = -∞.

You must write an algorithm that runs in O(log n) time.

 제한 사항

• 1 <= nums.length <= 1000
• -231 <= nums[i] <= 231 - 1
• nums[i] != nums[i + 1] for all valid i.

 입출력 예

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.

 Python 코드

Python code 

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        left = 0
        right = len(nums)-1
				# left가 커지고 right가 작아지는 경우 left와 right가 같아질 수 있다.
        # 그 때 원한느 값을 찾은 경우이므로 left와 right가 같아지면 반복문 종료
        while left < right:
            mid = (left + right) // 2      # 가운데 지점
            if nums[mid] < nums[mid+1]:    # 현재 값이 오른쪽 값보다 작다면
                left = mid+1
            else:                          # 현재 값이 오른쪽 값보다 크다면
                right = mid
        return left                        # 반복문이 종료되면 left가 정답

* 참고 링크 : https://velog.io/@hrpp1300/LeetCode-162.-Find-Peak-Element

 C++ 코드

C ++ code

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int left = 0;
        int right = nums.size() - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] > nums[mid + 1]) {
                right = mid;
            }
            else {
                left = mid + 1;
            }
        }
        return left; 
    }
};

* 참고 링크 : https://myeongcs.tistory.com/80

 출처

https://leetcode.com/problems/find-peak-element/submissions/