162. Find Peak Element

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.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

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.

Solution

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37

class Solution {
    public int findPeakElement(int[] nums) {
        int lo = 0; 
        int hi = nums.length - 1;
        
        if (nums.length == 1) {
            return 0;
        }
        
        while (lo <= hi) {
            
            int mid = lo + (hi - lo) / 2;
            
            if (isPeak(nums, mid)) {
                return mid;
            } else if (nums[mid + 1] > nums[mid]) {
                lo = mid + 1;
            } else {
                hi = mid - 1;
            }
            
        }
        return lo;
    }
    
    boolean isPeak(int[] nums, int x) {
        if (x == 0) {
            return nums[x] > nums[x + 1];
        }
        
        if (x == nums.length - 1) {
            return nums[x] > nums[x - 1];
        }
        
        return nums[x] > nums[x - 1] && nums[x] > nums[x + 1];
    }
}

Solution II

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16

class Solution {
    public int findPeakElement(int[] nums) {
      int lo = 0;
      int hi = nums.length - 1;
        while (lo < hi) {
            int mid = lo + (hi - lo) / 2;
            if (nums[mid] < nums[mid + 1]) {
                lo = mid + 1;
            } else {
                hi = mid;
            }
        }
        return lo;
    }

}