[LeetCode] 162. Find Peak Element

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mb5ff592736e0cf 2019-11-04 12:41:00

文章标签 leetcode binary search java javascript 数组 文章分类 后端开发

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
-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
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

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

时间O(logn)

空间O(1)

JavaScript实现

 1 /**
 2  * @param {number[]} nums
 3  * @return {number}
 4  */
 5 var findPeakElement = function(nums) {
 6     let start = 0;
 7     let end = nums.length - 1;
 8     while (start + 1 < end) {
 9         let mid = Math.floor(start + (end - start) / 2);
10         if (nums[mid] > nums[mid + 1]) {
11             end = mid;
12         } else {
13             start = mid;
14         }
15     }
16     if (nums[start] > nums[end]) return start;
17     return end;
18 };

Java实现

 1 class Solution {
 2     public int findPeakElement(int[] nums) {
 3         int start = 0;
 4         int end = nums.length - 1;
 5         while (start + 1 < end) {
 6             int mid = start + (end - start) / 2;
 7             if (nums[mid] > nums[mid + 1]) {
 8                 end = mid;
 9             } else {
10                 start = mid + 1;
11             }
12         }
13         if (nums[start] > nums[end]) {
14             return start;
15         } else {
16             return end;
17         }
18     }
19 }