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.
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
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
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;
}
}