leetcode 357 Count Numbers with Unique Digits

问题描述：
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])

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Following the hint. Let f(n) = count of number with unique digits of length n.

f(1) = 10. (0, 1, 2, 3, …., 9)

f(2) = 9 * 9. Because for each number i from 1, …, 9, we can pick j to form a 2-digit number ij and there are 9 numbers that are different from i for j to choose from.

f(3) = f(2) * 8 = 9 * 9 * 8. Because for each number with unique digits of length 2, say ij, we can pick k to form a 3 digit number ijk and there are 8 numbers that are different from i and j for k to choose from.

Similarly f(4) = f(3) * 7 = 9 * 9 * 8 * 7….

…

f(10) = 9 * 9 * 8 * 7 * 6 * … * 1

f(11) = 0 = f(12) = f(13)….

any number with length > 10 couldn’t be unique digits number.

The problem is asking for numbers from 0 to 10^n. Hence return f(1) + f(2) + .. + f(n)

As @4acreg suggests, There are only 11 different ans. You can create a lookup table for it. This problem is O(1) in essence.

  public int countNumbersWithUniqueDigits(int n) {        if (n == 0)     return 1;        int res = 10;        int uniqueDigits = 9;        int availableNumber = 9;        while (n-- > 1 && availableNumber > 0) {            uniqueDigits = uniqueDigits * availableNumber;            res += uniqueDigits;            availableNumber--;        }        return res;    }