Question

我有一个类似下图所示的循环。我有兴趣为这个循环结构找到Big O.

for i = 1 to n { // assume that n is input size
                         ...
                         for j = 1 to 2 * i {
                             ...
                             k = j;
                                 while (k >= 0) {
                                     ...
                                     k = k - 1;
                                 }
                         }
               }

我可以收集的是：

最外面的循环'n'次
第二个循环运行'2n'次（假设增量大小为1）
最内圈循环'2n'次

那么n的大O应该是O（n ^ 3）还是会有不同的东西呢？

将会对这些问题及其解决方案的具体链接表示赞赏。

Answer 1

让I（），M（），T（）成为内循环，中循环和整个程序（最外循环）的运行时间。如果我们从内到外工作，我们得到：

Inner-most loop;
I(j) = Summation (1) //from k=0 to j
I(j) = j+1  //Using basic Summation expansion formula.

Middle loop
M(i) = Summation (I(j)) //from j=1 to 2i
M(i) = Summation (j+1)  //from j=1 to j=2i with I(j)'s values
M(i) = Summation (j) + Summation (1)  //both from j=1 to j=2i

Using the expansion formula for Summation (j) from j=1 to n is '(n(n+1)/2)' and the fact that Summation (1) from j=1 to n is 'n', we get:

M(i) = 2i^2 + 3i

Outer-most loop:

T(n) = Summation (2i^2 + 3i)  //Summation from i=1 to n
T(n) = Summation (2i^2) + Summation (3i)  //both summations from i=1 to n
T(n) = 2*Summation (2i^2) + 3*Summation (i)  //both summations from i=1 to n
T(n) = (2(2n^3 + 3n^2 + n))/6) + (3(n(n+1))/2)  //using summation expansion formulas
T(n) = (4n^3 + 15n^2 + 11n)/2

Which means Big O of n be T(n^3).

注意：汇总扩展的基本求和扩展公式可以在this链接的第一页找到

感谢提示@Paul Hankin

找到嵌套循环结构的大O.

1 个答案: