leetcode 之 Count Numbers with Unique Digits

题目描述Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.

Example:
 Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99]) 

Hint:
1.A direct way is to use the backtracking approach.
2.Backtracking should contains three states which are (the current number, 
number of steps to get that number and a bitmask which represent which number is marked as visited so far in the current number). 
Start with state (0,0,0) and count all valid number till we reach number of steps equals to 10n.
3.This problem can also be solved using a dynamic programming approach and some knowledge of combinatorics.
4.Let f(k) = count of numbers with unique digits with length equals k.
5.f(1) = 10, ..., f(k) = 9 * 9 * 8 * ... (9 - k + 2) [The first factor is 9 because a number cannot start with 0].

即给定一个整数n, 计算从[0, 10^n)中,所有不含重复数字的数的个数

1. 回溯:

输入n的话, 不考虑10^n, 则总共有n位数. 使用 cur[n] 代表当前的数字. 使用一个flag[10] 代表0~9位数字,  每次放入一个数字,则flag对应的数字置为1,

代码如下:

class Solution(object):
    def countNumbersWithUniqueDigits(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n == 0:
            return 1
        flag = [0]*10
        cur = [0]*n
        step = 0;
        count = [0];
        self.helper(0, flag, cur, count, n)
        return count[0] 
    def helper(self, step, flag, cur, count, n):
        if step >= n:
            count[0] += 1
            return
        for i in range(0, 10):
            if flag[i] != 1 or (i==0 and step !=n and sum(cur[step:])==0):  #解决诸如[0,0,0], [0, 1,0]这类数字不能统计的问题
                cur[step] = i
                flag[i] = 1
                self.helper(step+1, flag, cur, count, n)
                flag[i] = 0
            else:
                continue

上述回溯法,在时间上超时了T_T

2.动态规划:

根据提示4,  f(k) = 9 * 9 *...(9-k+2)

class Solution(object):
    def countNumbersWithUniqueDigits(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n == 0:
            return 1
        if n == 1:
            return 10
        res = 0
        for i in range (1, n+1):
            res += self.count(i)
        return res
    def count(self, num):
        if num == 0 : return 1
        if num == 1: return 10
        res = 1
        for i in range(9, 11-num-1, -1):
            res *= i
        return res*9