伯克利CS61A错题笔记1.0

前段时间刷着berkeley的CS61A，在第二次作业中遇到一个题，魔怔了好久，解出来才发现根本不难，笔记纪念一下。

倒序比数，数大即停

原题 —— Count Until Larger

Implement the function count_until_larger. count_until_larger takes in a positive integer num. count_until_larger counts the distance between the rightmost digit of num and the nearest greater digit; to do so, the function counts digits from right to left. Once it encounters a digit larger than the rightmost digit, it returns that count. If no such digit exists, then the function returns -1.

For example, 8117 has a rightmost digit of 7 and returns a count of 3.
9118117 also returns a count of 3: for both, the count stops at 8.
0 should be treated as having no digits and returns a count of -1.

def count_until_larger(num):"""Complete the function count_until_larger that takes in a positive integer num.count_until_larger examines the rightmost digit and counts digits from right toleft until it encounters a digit larger than the rightmost digit, then returns that count.>>> count_until_larger(117) # .Case 1-1>>> count_until_larger(8117) # .Case 23>>> count_until_larger(9118117) # .Case 33>>> count_until_larger(8777)  # .Case 43>>> count_until_larger(22) # .Case 5-1>>> count_until_larger(0) # .Case 6-1""""*** YOUR CODE HERE ***"

题目大意就是设计一个函数 count_until_larger：输入任意自然数，找到【比最后一位数大、且与最后一位距离最近的】数，返回它与最后一位数的距离。如果输入是0或者找不到此数，则返回-1。

大致的思路：

倒序取数，循环比较最后一位和往前数字的大小，
往前数字大于最后一位时：停止，return 循环的次数
所有数字循环完成，未找到更大数字：return -1

第一版答案，用string非常暴力拆解了数字：

def count_until_larger(num):rightmost,listn,count = num%10,[],0listn.extend(i for i in str(num//10))listn.reverse()for number,i in enumerate(listn):if int(i)> rightmost:return number+1else:count += 1if count != 0:return -1else:return count

第二版优化后的答案，用取余数的方法一步步拆解数字：

def count_until_larger(num):rightmost, aheadnum, count = num%10,((num - num%10)%10), 0while num > 1:if aheadnum > rightmost:return countelse:count += 1num = num // 10aheadnum = num %10return -1

第二版其实和标答已经非常相似了，但也还是有值得优化的：

前一位数可以直接用 num %10 （除10、取余数）表示
当 num 为个位数时，num //10 = 0 （取整除结果），所以不需要 num > 1

附上标答：

def count_until_larger(num):rightmost = num % 10count = 0while num:if num % 10 > rightmost:return countnum = num // 10count = count + 1return -1

折腾了一中午，鬼知道一开始看这题的时候陷入了什么bug中，解决了也是很爽～