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PeakElement.java
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PeakElement.java
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package Leetcode;
/**
* @author kalpak
*
* 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] = -∞.
*
* 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.
*
*
* Follow up: Could you implement a solution with logarithmic complexity?
*/
public class PeakElement {
public static int findPeakElement(int[] nums) {
if (nums.length == 1)
return 0;
int left = 0;
int right = nums.length - 1;
while(left <= right) {
int mid = left + (right - left)/2;
if(mid > 0 && mid < nums.length - 1) {
if(nums[mid] > nums[mid - 1] && nums[mid] > nums[mid + 1])
return mid;
else if (nums[mid - 1] > nums[mid])
right = mid - 1;
else
left = mid + 1;
} else if (mid == 0) { // element on the left edge
if(nums[mid] > nums[mid + 1])
return 0;
else
return 1;
} else if (mid == nums.length - 1) { // element on the right edge
if(nums[mid] > nums[mid - 1])
return nums.length - 1;
else
return nums.length - 2;
}
}
return -1;
}
public static void main(String[] args) {
System.out.println("Index of Peak Element: " + findPeakElement(new int[]{1,2,1,3,5,6,4}));
}
}