357. Count Numbers with Unique Digits

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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])

解法1：

一个排列组合的题目，注意一位的数有10种选择，而大于一位的数第一位只有9种。
用一个数组或者一个变量prev记录之前的结果简化运算。
C++

1

Java

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public class Solution {
    public int countNumbersWithUniqueDigits(int n) {
        if (n == 0) {
            return 1;
        }
        if (n == 1) {
            return 10;
        }
        int sum = 0;
        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = 10;
        dp[2] = 81;
        sum += dp[1] + dp[2];
        for (int i = 3; i <= n; i++) {
            dp[i] = dp[i - 1] * (9 - i + 2);           
            sum += dp[i];
        }
        return sum;
    }
}