鉴于。。
T(0) = 3 for n <= 1
T(n) = 3T(n/3) + n/3 for n > 1
所以答案应该是
O(nlogn)
我是这样做的,但它没有给我正确的答案:T(n) = 3T(n/3) + n/3
T(n/3) = 3T(n/3^2) + n/3^2
把这个代入t(n)给出..
T(n) = 3(3T(n/3^2) + n/3^2) + n/3
T(n/3^2) = 3(3(3T(n/3^3) + n/3^3) + n/3^2) + n/3
最终看起来…
T(n) = 3^k (T(n/3^k)) + cn/3^k
设置
k = lgn..
T(n) = 3^lgn * (T(n/3^lgn)) + cn/3^lgn
T(n) = n * T(0) + c
T(n) = 3n + c
但答案是……我的步骤有什么问题?
最佳答案
T(n) = 3T(n/3) + n/3
T(n/3) = 3T(n/9) + n/9
T(n) = 3(3T(n/9) + n/9) + n/3
= 9T(n/9) + 2*n/3 //statement 1
T(n/9)= 3T(n/27) + n/27
T(n) = 9 (3T(n/27)+n/27) + 2*n/3 // replacing T(n/9) in statement 1
= 27 T (n/27) + 3*(n/3)
T(n) = 3^k* T(n/3^k) + k* (n/3) // eventually
用log n替换k到基3。
T(n) = n T(1) + (log n) (n/3);
// T(1) = 3
T(n) = 3*n + (log n) (n/3);
Hence , O (n* logn)
关于algorithm - 解决类似的重复:T(n)= 3T(n/3)+ n/3,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/23259404/