LeetCode 357. Count Numbers with Unique Digits（计算无重复数字）

原题网址：https://leetcode.com/problems/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])

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].

方法：动态规划。对数据样例进行分析，主要是如何解决0的问题，0可以出现在除了最高位之外的任何位置（除非0数值本身），这样最高位总是有1~9的可能性。

public class Solution {
    public int countNumbersWithUniqueDigits(int n) {
        if (n < 0) return 0;
        if (n == 0) return 1;
        if (n == 1) return 10;
        int counts = 10;
        int m = 9;
        for(int i=1, j=9; i0; i++, j--) {
            m *= j;
            counts += m;
        }
        return counts;
    }
}