题目描述
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)
- 单调递减,这种情况下,
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:
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