做个简单记录, 如有疏漏, 欢迎指正
第一种是时间复杂度为 2^n 的递归实现
1 size_t rec_fibonacci(int idx) 2 { 3 if (0 >= idx) return 0; 4 if (2 >= idx) return idx; 5 6 return (rec_fibonacci(idx - 1) + rec_fibonacci(idx - 2)); 7 }
第二种是时间复杂度为 n 的类似动态规划实现
1 size_t dp_fibonacci(int idx) 2 { 3 if (0 >= idx) return 0; 4 if (2 >= idx) return idx; 5 6 std::vector<size_t> fibonacci; 7 fibonacci.push_back(1); 8 fibonacci.push_back(2); 9 10 size_t pre_last_num, last_num, curr_num; 11 12 pre_last_num = fibonacci[0]; 13 last_num = fibonacci[1]; 14 15 for (size_t i = 2; i < idx; ++i) 16 { 17 curr_num = pre_last_num + last_num; 18 pre_last_num = last_num; 19 last_num = curr_num; 20 21 fibonacci.push_back(curr_num); 22 } 23 24 return fibonacci[idx - 1]; 25 }