143.Count Numbers with Unique Digits

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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].

 /*** 对于k位数，统计出位数小于等于k的各个数字位上面数字均不相同的总个数。* 例如：k=1时，结果为10;k=2时，结果为91;k=3时，结果为739.* k=3时，三位数则根据排列计算9*9*8=648(首位是9是因为三位数的最高位不可为0)* 两位数则9*9=81，一位数为10,648+81+10=739.* * 令f(k)表示给定数字k时的结果，则f(k) = f(k-1) + 9*9*8***(9-k+2)* 故可采用动态规划的思想。* 如果是十一位数及以上，则符合各个位上数字不同的整数个数为0.**/public int CountNumbersWithUniqueDigits(int n){if(n == 0){return 1;}if(n == 1){return 10;}int f_k = 1;int f_k1 = 10;int pro = 9;//保存的是上一次计算的排列的结果for (int i = 2; i<=10 && i <= n;i++ ){pro = pro * (11 - i);f_k = f_k1 + pro;f_k1 = f_k;}return f_k;}

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