看《面试宝典》遇到一道题才知道T(n),一直只用O(n)表示时间复杂度了。
时间复杂度的定义
一般情况下,算法中基本操作重复执行的次数是问题规模n的某个函数,用T(n)表示,若有某个辅助函数f(n),使得当n趋近于无穷大时,T(n)/f(n)的极限值为不等于零的常数,则称f(n)是T(n)的同数量级函数。记作T(n)=O(f(n)),称O(f(n))为算法的渐进时间复杂度(O是数量级的符号 ),简称时间复杂度。
对于T(n) = a*T(n/b)+c*n^k;T(1) = c 这样的递归关系,有这样的结论:
if (a > b^k) T(n) = O(n^(logb(a)));logb(a)b为底a的对数
if (a = b^k) T(n) = O(n^k*logn);
if (a < b^k) T(n) = O(n^k);
a=25; b = 5 ; k=2
a==b^k 故T(n)=O(n^k*logn)=O(n^2*logn)
T(n) = 25T(n/5)+n^2
= 25(25T(n/25)+n^2/25)+n^2
= 625T(n/25)+n^2+n^2 = 625T(n/25) + 2n^2
= 25^2 * T( n/ ( 5^2 ) ) + 2 * n*n
= 625(25T(n/125)+n^2/625) + 2n^2
= 625*25*T(n/125) + 3n^2
= 25^3 * T( n/ ( 5^3 ) ) + 3 * n*n
= ....
= 25 ^ x * T( n / 5^x ) + x * n^2
T(n) = 25T(n/5)+n^2
T(0) = 25T(0) + n^2 ==> T(0) = 0
T(1) = 25T(0)+n^2 ==> T(1) = 1
x = lg 5 n
25 ^ x * T( n / 5^x ) + x * n^2
= n^2 * 1 + lg 5 n * n^2
= n^2*(lgn)
= 625T(n/25)+n^2+n^2 = 625T(n/25) + 2n^2
= 25^2 * T( n/ ( 5^2 ) ) + 2 * n*n
= 625(25T(n/125)+n^2/625) + 2n^2
= 625*25*T(n/125) + 3n^2
= 25^3 * T( n/ ( 5^3 ) ) + 3 * n*n
= ....
= 25 ^ x * T( n / 5^x ) + x * n^2
T(n) = 25T(n/5)+n^2
T(0) = 25T(0) + n^2 ==> T(0) = 0
T(1) = 25T(0)+n^2 ==> T(1) = 1
x = lg 5 n
25 ^ x * T( n / 5^x ) + x * n^2
= n^2 * 1 + lg 5 n * n^2
= n^2*(lgn)