[leetcode] 357. Count Numbers with Unique Digits 解题报告

题目链接：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].

思路：一个排列组合的题目，让求没有重复数字的数的个数，学过概率论的应该很容理解．当n=1时因为只有一个数字，所以0-9都是答案．当n>=2时，最高位可以为1-9任意一个数字，之后各位可以选择的数字个数依次为9, 8, 7, 6...，上一位选一个下一位就少了一种选择．

代码如下：

class Solution {
public:
    int countNumbersWithUniqueDigits(int n) {
        if(n==0) return 1;
        if(n==1) return 10;
        int val = 9, ans = 10;
        for(int i = 2; i <= n; i++)
        {
            val *= (9-i+2);
            ans += val;
        }
        return ans;
    }
};