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# Solution 1 bisearch

* t-complexity $O(logn)$
* s-complexity $O(1)$

O(logN)一般考虑二分搜索。有如下规律：

规律一：如果nums\[i+1] < nums\[i]，则在i之前一定存在峰值元素（包括i）

规律二：如果nums\[i] < nums\[i+1]，则在i+1之后一定存在峰值元素（包括i+1）如果i+1是最后一个元素，则i+1就是峰值