leetcode 357. Count Numbers with Unique Digits

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])

x<10n 就是有n位去放数，而且放的数字不能相同，要注意的是最高位为0是不算数字的，也就是说高位可以有多个0

很好的解释

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)

代码也很简单

public class Solution {
    public int countNumbersWithUniqueDigits(int n) {
        if(n==0) return 1;

        int res = 10;
        int cur = 9;
        int t = 10;
        while(n-->1 && t-- >=0){
            cur *=t;
            res += cur;
        }
        return res;
    }
}