下面是一个递归 O（log n）的算法。

假设我们有数字数组，而我们知道，该段的中间数不大于端点小

Suppose we have an array of numbers, and we know that the middle number of that segment is no smaller than the endpoints:

A[i] <= A[m] >= A[j]

对于I，J指标到一个数组，而 M =（I + J）/ 2 。检查中途端点和中点之间的所有元素，即那些在指数 X =（3 * I + J）/ 4 和 Y =（我+ 3 * j）条/ 4 。如果 A [X]＆GT; = A [M] ，然后递归的间隔 [I，M] 。如果 A [Y]＆GT; = A [M] ，然后递归的间隔 [M，J] 。否则，递归在区间 [X，Y] 。

for i,j indexes into an array, and m=(i+j)/2. Examine the elements midway between the endpoints and the midpoint, i.e. those at indexes x=(3*i+j)/4 and y=(i+3*j)/4. If A[x]>=A[m], then recurse on the interval [i,m]. If A[y]>=A[m], then recurse on the interval [m,j]. Otherwise, recurse on the interval [x,y].

在每一种情况下，我们维持上面的间隔不变。最终我们得到大小为2的区间，这意味着我们已经找到了一个峰值（即 A [M] ）。

In every case, we maintain the invariant on the interval above. Eventually we get to an interval of size 2 which means we've found a peak (which will be A[m]).

要圆转换为一个阵列，取3等距​​离的样品和确定自己的方位，以使最大（或一个并列最大）为在区间的中部，另两个点的端点。运行时间为 O（log n）的，因为每个间隔的previous一半大小。

To convert the circle to an array, take 3 equidistant samples and orient yourself so that the largest (or one tied for the largest) is in the middle of the interval and the other two points are the endpoints. The running time is O(log n) because each interval is half the size of the previous one.

我已经掩盖了如何计算索引时四舍五入问题，但我认为你可以工作了这一点成功。

I've glossed over the problem of how to round when computing the indexes, but I think you could work that out successfully.