[LeetCode] 162. Find Peak Element
A peak element is an element that is strictly greater than its neighbors.
Given a 0-indexed 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] = -∞. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.
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
-231 <= nums[i] <= 231 - 1
nums[i] != nums[i + 1] for all valid i.
寻找峰值。
峰值元素是指其值大于左右相邻值的元素。
给你一个输入数组 nums,找到峰值元素并返回其索引。数组可能包含多个峰值,在这种情况下,返回 任何一个峰值 所在位置即可。
你可以假设 nums[-1] = nums[n] = -∞ 。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/find-peak-element
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思路
给一个数组,数组满足条件nums[i] ≠ nums[i+1],求数组峰值的下标。这个题的👎多于👍,估计是因为峰值不止一个吧,我做的时候也踩了坑。
思路是用二分法,因为题目要求时间复杂度是 log 级别。根据左右指针计算中间位置 m,并比较 m 与 m+1 的值,如果 m 较大,则左侧存在峰值,r = m,如果 m + 1 较大,则右侧存在峰值,l = m + 1。
复杂度
时间O(logn)
空间O(1)
代码
Java实现
class Solution {
public int findPeakElement(int[] nums) {
int start = 0;
int end = nums.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
// 峰值可能在mid左侧
if (nums[mid] > nums[mid + 1]) {
end = mid;
}
// 峰值在mid右侧
else {
start = mid;
}
}
if (nums[start] > nums[end]) {
return start;
}
return end;
}
}
JavaScript实现
/**
* @param {number[]} nums
* @return {number}
*/
var findPeakElement = function (nums) {
let start = 0;
let end = nums.length - 1;
while (start + 1 < end) {
let mid = Math.floor(start + (end - start) / 2);
if (nums[mid] > nums[mid + 1]) {
end = mid;
} else {
start = mid;
}
}
if (nums[start] > nums[end]) {
return start;
}
return end;
};
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