打印1到最大的n位数 链接到标题 # 将数组转换为字符 def print_number(number): is_beginning_0 = True num_len = len(number) for i in range(num_len): if is_beginning_0 and number[i] != "0": is_beginning_0 = False if not is_beginning_0: print("%c" % number[i], end="") print("") # 边界条件：n > 0 def print_1_to_max_of_n1(n): if n <= 0: return number = ["0"] * n while not increment(number): print_number(number) # 从最后一位开始计算，如果 最后一位增长为10，则重置为 0，且进位；如果首位增长为 10，则溢出 def increment(number): is_carry = 0 is_overflow = False sum = 0 num_len = len(number) for i in range(num_len - 1, -1, -1): sum = int(number[i]) + is_carry if i == num_len - 1: sum += 1 if sum >= 10: if i == 0: is_overflow = True else: sum -= 10 number[i] = str(sum) is_carry = 1 else: number[i] = str(sum) break return is_overflow def print_1_to_max_of_n2(n): if n <= 0: return number = ["0"] * n for i in range(10): number[0] = str(i) print_1_to_max_of_n_recursively(number, n, 0) # 始终找到最后一位，并将其计算 def print_1_to_max_of_n_recursively(number, num_len, index): if index == num_len - 1: print_number(number) return for i in range(10): number[index + 1] = str(i) print_1_to_max_of_n_recursively(number, num_len, index +1) print_number(["0", "1", "1"]) print_1_to_max_of_n1(2) print_1_to_max_of_n2(2)