https://leetcode.com/problems/find-peak-element/

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.

程式碼 1：

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int n = nums.size();
        if(n == 1) return 0;
        if(n == 2) {
            if(nums[1] > nums[0]) return 1;
            else return 0;
        }
        vector<int> ans;
        int maxx = nums[0];
        for(int i = 1; i < n - 1; i ++) {
            int temp = i;
            //if(nums[temp] < maxx) return 0;
            while(nums[temp] > maxx && temp < n) {
                maxx = nums[temp];
                temp ++;
            }
            return temp - 1;
        }
        return -1;
    }
};

　　寫的亂糟糟的一個程式碼 我可能是豬腦子吧 只要找出愛第一個後一個比前一個小的然後返回就好了 否則就是一個排好序的陣列返回最後一個的下標是最大值就好了 

程式碼 2：

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        for (int i = 1; i < nums.size(); ++i) {
            if (nums[i] < nums[i - 1]) return i - 1;
        }
        return nums.size() - 1;
    }
};