小于n的最大数 Leetcode 902 Numbers At Most N Given Digit Set

这两个问题的本质就是一个棵树，然后根据n对树做剪枝。难点在于剪的时候边界条件有些坑，get_lower_largest_digit_dic是这两个题目的共同点

题目一： 小于n的最大数

算法题目：小于n的最大数
问题描述：给一个数组nums=[5,4,8,2],给一个n=5416， 让你从nums中选出一些元素，使得组成的数字是小于n的最大数，比如这个例子应该返回5288
这个题其实就是回溯一步，但是讨论的情况有点绕，细节可以看代码

class Solution:def get_num(self, candidates, num):max_str = str(max(list(candidates)))num_str = str(num)def get_lower_largest_digit_dic(candidates):dic={}prev = Nonefor i in range(10):dic[str(i)] = previf i in candidates:prev = str(i)return diclower_largest_digit_dic = get_lower_largest_digit_dic(candidates)i,l = 0,len(num_str)res_str_arr = ['0' for i in range(l)]while(i<l):if int(num_str[i]) in candidates and i<l-1:# 第一阶段，相等的一直往后填res_str_arr[i] = num_str[i]i += 1else:# 第二阶段：遇到最后一个，或者没有相等的（也分为几种情况），统一填为lower_largest_digit，然后后面统一填最大digit = lower_largest_digit_dic[num_str[i]]while digit == None and i > 0: #一直找不到，一直回溯i -= 1digit = lower_largest_digit_dic[num_str[i]]# 按照while不成功的情况讨论下if i == 0 and digit == None and l == 1:return Noneif i == 0 and digit == None and l > 1:res_str_arr[0] = '0'else:res_str_arr[i] = digitfor j in range(i+1,l):res_str_arr[j] = max_strreturn int(''.join(res_str_arr))if __name__ == '__main__':s = Solution()print(s.get_num({1,2,9,4}, 2533))  # 2499print(s.get_num({1,2,5,4}, 2543))  # 2542print(s.get_num({1,2,5,4}, 2541))  # 2525print(s.get_num({1,2,9,4}, 2111))  # 1999print(s.get_num({5,9}, 5555)) #999

题目二： 最大为 N 的数字组合

来自https://leetcode.com/problems/numbers-at-most-n-given-digit-set/

Given an array of digits which is sorted in non-decreasing order. You can write numbers using each digits[i] as many times as we want. For example, if digits = [‘1’,‘3’,‘5’], we may write numbers such as ‘13’, ‘551’, and ‘1351315’.

Return the number of positive integers that can be generated that are less than or equal to a given integer n.

Example 1:

Input: digits = [“1”,“3”,“5”,“7”], n = 100
Output: 20
Explanation:
The 20 numbers that can be written are:
1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.
Example 2:

Input: digits = [“1”,“4”,“9”], n = 1000000000
Output: 29523
Explanation:
We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers,
81 four digit numbers, 243 five digit numbers, 729 six digit numbers,
2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers.
In total, this is 29523 integers that can be written using the digits array.
Example 3:

Input: digits = [“7”], n = 8
Output: 1

Constraints:

1 <= digits.length <= 9
digits[i].length == 1
digits[i] is a digit from ‘1’ to ‘9’.
All the values in digits are unique.
digits is sorted in non-decreasing order.
1 <= n <= 109

其实和题目一思路很相似，但是边界条件容易错：

class Solution:def atMostNGivenDigitSet(self, digits: List[str], n: int) -> int:def get_lt_digit_cnt_dic(digits_set):dic = {}prev = 0for i in range(0,10):stri = str(i)dic[stri] = previf stri in digits_set:prev += 1return dicdigits_set = set(digits)lt_digit_cnt_dic = get_lt_digit_cnt_dic(digits_set)dl = len(digits)nl = len(str(n))factorial = [1 for i in range(nl)]for i in range(1,nl):factorial[i] = factorial[i-1]*dlres1 = sum(factorial[i] for i in range(1,nl))res2,i = 0,0strn = str(n)while i<nl:lt_digit_cnt = lt_digit_cnt_dic[strn[i]]res2 += lt_digit_cnt * factorial[nl-(i+1)]if (strn[i] not in digits_set):breaki+=1# 这个条件容易漏掉，例如digits =["3","4","8"], n = 4if i == nl and strn[nl-1] in digits_set:res2 += 1# print(res1)return res1+res2