LeetCode #162 - Find Peak Element-优快云博客

本文介绍了一种在给定数组中寻找峰值元素的方法，并确保了算法的时间复杂度为O(logn)。峰值元素定义为一个大于其左右邻居的元素，通过使用二分查找策略，有效地定位到了峰值位置。

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题目描述：

A peak element is an element that is greater than its neighbors.

Given an input array nums, where nums[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 that nums[-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.

给定一个数组，相邻的两个数不相等，找到一个数，使它的值大于左右邻居，而且时间复杂度必须为O(log n)。由于只需要找到一个局部峰值，所以可以采用二分法，确定好中间元素后，把它和后一个元素比较，如果较大说明峰值在前半段，如果较小说明峰值在后半段。

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int begin=0;
        int end=nums.size()-1;
        while(begin<=end)
        {
            int mid=(begin+end)/2;
            if(begin==end) return mid;
            else
            {
                if(nums[mid]<nums[mid+1]) begin=mid+1;
                else end=mid;
            }
        }
    }
};