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對(duì)于斐波那契數(shù),若是采用遞歸的算法,每個(gè)遞歸調(diào)用都將觸發(fā)另外兩個(gè)遞歸調(diào)用,而這兩個(gè)中調(diào)用任意一個(gè)還會(huì)觸發(fā)另外兩個(gè)的調(diào)用。遞歸調(diào)用的時(shí)間復(fù)雜度O(2^N),空間復(fù)雜度為O(N),所以在計(jì)算略大的數(shù)會(huì)花費(fèi)一定的時(shí)間和空間。遞歸程序如下:
#include<iostream> using namespace std; unsigned long long Fib(size_t num) { if (num < 2) { return num; } else return Fib(num - 1) + Fib(num - 2); } int main() { unsigned long long ret = Fib(10); cout << ret << endl; system("pause"); return 0; }
用迭代方法計(jì)算第N 個(gè)斐波那契數(shù),時(shí)間復(fù)雜度O(N),空間復(fù)雜度O(1),程序如下:
#include<iostream> using namespace std; unsigned long long Fib(size_t num) { unsigned long long first = 0; unsigned long long second = 1; unsigned long long sum = 0; if (num < 2) return num; else for (size_t i = 2; i <= num; i++) { sum = first + second; first = second; second = sum; } return sum; } int main() { unsigned long long ret = Fib(10); cout <<"Fibonacci(10)="<< ret << endl; system("pause"); return 0; }
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