原文地址：T(n) = 25T(n/5)+n^2的时间复杂度 作者：djkpengjun 主定理： 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)
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