357. Count Numbers with Unique Digits

题目

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].

分析

从最高位开始，分析每位可能的数字个数，如果n=1，有10种可能，n=2，则十位有9种（除去0），个位有9种（除去十位的选择，加上0），以此类推，得到f(k)函数，计算k位情况下每位不重复的数字个数，则在 0 ≤ x < 10n范围之内的数字个数为f(1)+f(2)+...+f(n)。

class Solution {public:    int countNumbersWithUniqueDigits(int n) {        if(n==0)            return 1;        int countNumber=10;//计算总数，初始化为10        int backtrack=9;//计算f(k)的前k-1项乘积，从第二位开始计算，故初始化为9        if(n>10)//当n>10时，无法再构造更多的不重复数字，故大于10位都当做10位算            n=10;        for(int i=2;i<=n;++i)//从第二位开始计算f(k)        {            backtrack*=9-i+2;//计算f(k)，得到k位不重复数字的数目            countNumber+=backtrack;//与总数相加        }        return countNumber;            }};