A peak element is an element that is greater than its neighbors.
Given an input arraynums, wherenums[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 thatnums[-1] = nums[n] = -∞.
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: 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.
Solution
Linear solution that check whether the next element is smaller, when it is, peak was found.
public class Solution {
public int findPeakElement(int[] nums) {
for (int i = 0; i < nums.length - 1; i++) {
if (nums[i] > nums[i + 1])
return i;
}
return nums.length - 1;
}
}
Iterative binary search.
class Solution {
public int findPeakElement(int[] nums) {
int start = 0;
int end = nums.length - 1;
while (start < end) {
int middle = start + ((end - start) / 2);
int middleVal = nums[middle];
// 1, 2, 3, 1
if (middleVal > nums[middle + 1]) {
end = middle;
} else {
start = middle + 1;
}
}
return start;
}
}
Recursive binary search.
public class Solution {
public int findPeakElement(int[] nums) {
return search(nums, 0, nums.length - 1);
}
public int search(int[] nums, int l, int r) {
if (l == r)
return l;
int mid = (l + r) / 2;
if (nums[mid] > nums[mid + 1])
return search(nums, l, mid);
return search(nums, mid + 1, r);
}
}