162. Find Peak Element

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

Given a 0-indexed 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] = -∞. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.

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.

注意：

        1.明确一点，当middle不是峰值时，无论左边还是都应该会有一个峰值

        2.这里用到了一个lambda表达式，用来处理边界的情况，因为这里要比左右两边都大，要处理刚好在最左边和最右边的情况

        3.二分查找法：（1）当middle为峰值时，更新ret的值，并break，最后返回

                                 （2）如果middle对应的值比middle+1对应的大，就往右，否则往左。

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        auto get=[&](int i)->pair<int,int>{
            if(i==-1 || i==nums.size()){
                return {0,0};
            }
            return {1,nums[i]};
        };
        int left=0;
        int right=nums.size()-1;
        int ret=-1;
        while(left<=right){
            int middle=left+(right-left)/2;
            if(get(middle-1)<get(middle) && get(middle+1)<get(middle)){
                ret=middle;
                break;
            }else if(get(middle)>get(middle-1)){
                left=middle+1;
            }else{
                right=middle-1;
            }
        }
        return ret;
    }
};