LeetCode 357

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])

Credits:
Special thanks to @memoryless for adding this problem and creating all test cases.

Solution:

题意是返回0～10的n次方间，每个位数上的数字都不相同的数的总数。

根据不同的总位数n进行分析，num为位数为n时符合条件的数的总数。
n = 0时，num = 1;
n = 1时，num = 9;
n = 2时，num = 9*9；
n = 3时，num = 9*9*8;
….
n = 10时，num = 9*9*8*7*6*5*4*3*2*1;

hint：第一位有9个选择（除了0），第二位还剩9个选择（包括了0），第三位还有8个选择……

最后的返回值为所有可能的n的取值对应的num之和

class Solution {public:    int countNumbersWithUniqueDigits(int n) {        if(n == 0)            return 1;        if(n > 10)            n = 10;        int result = 1;        for(int i = 1; i <= n; i++)             result += getNum(i);        return result;    }    int getNum(int n) {        if(n == 0) {            return 1;        } else if(n == 1) {            return 9;        } else {            int result = 9;            for(int i = 0; i < n-1; i++)                 result = result*(9-i);            return result;        }    }};