摘要: # 递归。重复计算,时间复杂度O(2^n)- O(1) ,空间复杂度为O(n) def fRecursive(n): if n == 0: return 0 elif n == 1: return 1 else: return fRecursive(n - 1) + fRecursive(n - 2 阅读全文
posted @ 2020-06-28 21:08 江湖凶险 阅读(187) 评论(0) 推荐(0)