二分的运用：很是巧妙

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
      if (nums.size() == 1) {
        return 0;
      }
      
      int left = 0, right = nums.size() - 1;
      int mid = (left + right) / 2;
      
      // left 指向的值一定大于前一位
      // right指向的值一定大于后一位
      // 两者重合时就是峰值
      while (left < right) {
        if (nums[mid] < nums[mid + 1]) {
          left = mid + 1;
        } else {
          right = mid;
        }
        mid = (left + right) / 2;
      }
      
      return right;
    }
};