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[Leetcode]162. Find Peak Element(C++)

题目描述

题目链接: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] = -∞.

例子

例子 1

Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.

例子 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.

Follow up

Could you implement a solution with logarithmic complexity?

Constraints

  • 1 <= nums.length <= 1000
  • -2^31 <= nums[i] <= 2^31 - 1
  • nums[i] != nums[i + 1] for all valid i.

解题思路

这道题要求我们在无序数组中找到至少一个局部峰值(其值大于两侧元素的值)的下标。最简单的方法是直接遍历一次数组,逐个判断即可,时间复杂度为线性的。而题目中要求我们尝试在指数时间内找出来,因此可以从二分查找的思路来进行判断,首先确定左右边界,取中间值 Mid,进行以下判断:

  • nums[Mid] > nums[Mid + 1],此时对数组后半段 [Mid, N - 1] 有几种可能:
    • 单调递减,这种情况下,nums[N-1] 是峰值,因为题目假定 nums[N] = nums[-1] = -Inf
    • 非单调递减,那么至少会有一个峰值,因为 nums[Mid] > nums[Mid + 1],从 Mid + 1 往后遍历的第一个不是比后面元素大的即为局部峰值(注意:这里可以这么判断是因为题目给定了前后两个元素不能相等

结合以上情况,可以判断 [Mid, N - 1] 至少有一个峰值,因此收缩左边界: Left = Mid,反过来情况也是类似的,不过是将右边界收缩到 Right = Mid + 1

代码如下:

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class Solution {
public:
int findPeakElement(vector<int>& nums) {
int left = 0, right = nums.size() - 1;
while (left < right) {
int mid = (left + right) / 2;
if (nums[mid] > nums[mid + 1]) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
};
  • 时间复杂度: O(log n)
  • 空间复杂度: O(1)

GitHub 代码同步地址: 162.FindPeakElement.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions