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162. Find Peak Element
Description
A peak element is an element that is strictly greater than its neighbors.
Given a 0indexed 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 validi
.
Solutions
Solution 1: Binary Search
We define the left boundary of binary search as $left=0$ and the right boundary as $right=n1$, where $n$ is the length of the array. In each step of binary search, we find the middle element $mid$ of the current interval, and compare the values of $mid$ and its right neighbor $mid+1$:
 If the value of $mid$ is greater than the value of $mid+1$, there exists a peak element on the left side, and we update the right boundary $right$ to $mid$.
 Otherwise, there exists a peak element on the right side, and we update the left boundary $left$ to $mid+1$.
 Finally, when the left boundary $left$ is equal to the right boundary $right$, we have found the peak element of the array.
The time complexity is $O(\log n)$, where $n$ is the length of the array $nums$. Each step of binary search can reduce the search interval by half, so the time complexity is $O(\log n)$. The space complexity is $O(1)$.

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

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

class Solution: def findPeakElement(self, nums: List[int]) > int: left, right = 0, len(nums)  1 while left < right: mid = (left + right) >> 1 if nums[mid] > nums[mid + 1]: right = mid else: left = mid + 1 return left

func findPeakElement(nums []int) int { left, right := 0, len(nums)1 for left < right { mid := (left + right) >> 1 if nums[mid] > nums[mid+1] { right = mid } else { left = mid + 1 } } return left }

function findPeakElement(nums: number[]): number { let [left, right] = [0, nums.length  1]; while (left < right) { const mid = (left + right) >> 1; if (nums[mid] > nums[mid + 1]) { right = mid; } else { left = mid + 1; } } return left; }