LeetCode Count Numbers with Unique Digits(计数问题)

题意：给出一个整数，求在[0,10^n]之间各位数字不相同的数的个数

思路：n为1时，有10个

n为2时, 十位数只能是1-9,而个数位可以为0，所以也是有9种选择

n为3时，百位数有9种选择，十位数有9种选择，个数数有8种选择

依此类推

代码如下：

public class Solution
{public int countNumbersWithUniqueDigits(int n){if (0 == n) return 1;else if (1 == n) return 10;int sum = 10;int dp = 9;for (int i = 2; i <= n; i++){dp *= 11 - i;sum += dp;}return sum;}
}

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