Find Peak Element
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.
Analysis
乍一看此题中的数列并没有sorted过,是不是就不能用binary search了呢?
其实binary search最基本的特征其实在于根据判断条件,每次减半搜索空间(search space)。这里其实也可以,在于nums[mid]
可以与nums[mid+1]
比较大小,以此判断在上升趋势还是下降趋势,从而改变搜索边界。最终得到的index就(可能,取决于用哪个template)是局部的peak。
Solution
Template #3
class Solution {
public int findPeakElement(int[] nums) {
int left = 0, right = nums.length - 1;
while (left + 1< right) {
int mid = left + (right - left) / 2;
if (nums[mid] > nums[mid + 1]) {
// descending
right = mid;
} else {
// ascending
left = mid;
}
}
if (nums[left] >= nums[right]) {
return left;
}
return right;
}
}
Template #2
class Solution {
public int findPeakElement(int[] nums) {
int left = 0, right = nums.length - 1;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] > nums[mid + 1]) {
// descending
right = mid;
} else {
// ascending
left = mid + 1;
}
}
return left;
}
}
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