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] = -∞.

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.

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

• 一开始读错题了，以为输出的是元素，实际上要输出位置
• 之前写了前面的函数，内存占用太高了，改用while循环了
  想知道该怎样不每次都输入整个vector时也能用递归
  题解三种方法，在这里,前两种都是O(n)时间复杂度，第三种二分法和我的应该差不多，没细看，有一些没用过的操作需要看看