357. Count Numbers with Unique Digits

Description

Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10 n.

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].

Analysis

此题其实是一道非常容易理解的排列组合题，但由于题目描述过于庞大，往往会给人一种此题很难的感觉，下面用归纳推理的方式对此题进行解决:

当n==1时，由0~10之间的数字所组成的数字均符合题目要求，所以返回10

当n==2时，第一个位置上的数不能为0，所以有9种情况，对于第二个数，可以取0，所以也是9种情况，故有

9*9=81

种情况，再加上n==1时的10种情况，所以返回91

n==3及之后的情况请自行考虑，我们已经找出了解决此题的一种方法，代码如下:

Code

class Solution {
public:
    int countNumbersWithUniqueDigits(int n) {
        if(n==0)
            return 1;
        n=min(n,10);
        vector<int> dp(n+1,9);
        int sum=0;
        dp[1]=10;
        for(int i=2;i<=n;++i)
            for(int j=9;j>=9-i+2;--j)
                dp[i]*=j;
        for(int i=1;i<dp.size();++i){
            sum+=dp[i];
        }
        return sum;
    }
};