Algorithm
Problem Name: 162. Find Peak Element
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 <= 1000-231 <= nums[i] <= 231 - 1nums[i] != nums[i + 1]for all validi.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int findPeakElement(int* nums, int numsSize) {
#if 0
int i;
for (i = 1; i < numsSize && nums[i] > nums[i - 1]; i ++) {
}
return i - 1;
#else
int l, r, m;
if (numsSize == 1) return 0;
l = 0; r = numsSize - 1;
while (l < r) {
m = l + (r - l) / 2;
if (nums[m] < nums[m + 1]) {
l = m + 1;
} else {
r = m;
}
}
return l;
#endif
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int findPeakElement(vector<int>& nums) {
if (nums.size() == 1) {
return 0;
}
int n = nums.size();
if (nums[n - 2] < nums[n - 1]) {
return n - 1;
} else if (nums[1] < nums[0]) {
return 0;
}
int l = 1, r = n - 2, mid;
while (l < = r) {
mid = l + (r - l)/2;
if (nums[mid - 1] > nums[mid]) {
r = mid - 1;
} else if (nums[mid + 1] > nums[mid]) {
l = mid + 1;
} else {
return mid;
}
}
return -1;
}
};
Input
Output
#3 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int findPeakElement(int[] nums) {
return helper(nums, 0, nums.length - 1);
}
private int helper(int[] nums, int startIdx, int endIdx) {
if (startIdx == endIdx) {
return startIdx;
}
int midIdx = (startIdx + endIdx) / 2;
int nextToMid = midIdx + 1;
if (nums[midIdx] > nums[nextToMid]) {
return helper(nums, startIdx, midIdx);
} else {
return helper(nums, nextToMid, endIdx);
}
}
}
Input
Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const findPeakElement = function(nums) {
let low = 0;
let high = nums.length-1;
while(low < high) {
let mid1 = low + ((high - low) >> 1);
let mid2 = mid1 + 1;
if(nums[mid1] < nums[mid2]> low = mid2;
else high = mid1;
}
return low;
};
Input
Output
#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def findPeakElement(self, nums):
l, r, n = 0, len(nums) - 1, len(nums)
while l <= r:
mid = (l + r) // 2
pre, after = mid == 0 and -float("inf") or nums[mid - 1], mid == n - 1 and -float("inf") or nums[mid + 1]
if pre < nums[mid] > after: return mid
elif pre > nums[mid]: r = mid - 1
else: l = mid + 1
Input
Output
#6 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0162_FindPeakElement
{
public int FindPeakElement(int[] nums)
{
int left = 0, right = nums.Length - 1;
while (left < right)
{
var mid = left + (right - left) / 2;
if (nums[mid] > nums[mid + 1]) right = mid;
else left = mid + 1;
}
return left;
}
}
}
Input
Output