此代码计算序列 A 的子序列中存在的最小值。子序列从索引 i 开始,到索引 j 结束。您的算法可以用英文翻译为:
puzzle(A, i, j) :
if the subsequence has only one element :
return this element
min-left is the minimum value present at the first half of the subsequence(it's computed recursively)
min-right is the minimum value present at the second half of the subsequence(it's computed recursively)
return the minimum of min-left, min-right
这个算法的复杂度显然是线性的(O(N))。但是,如果你不相信我,我会用主定理(http://en.wikipedia.org/wiki/Master_theorem)为你证明。
让 T(N) 成为您的算法的递归。然后:
T(N) = 2T(N/2) =>
T(N) = 2T(N/2) + Theta(1) =>
T(N) = Theta(N) (from the first case of the master theorem, which states
that if f(N) = O(N^logb a-e), for e>0, then the complexity is Theta(N^logb a),
where a=2, b=2, f(N)=Theta(1), and e=1