LeetCode: 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]
)
题目意思是计算x中各个位不同的整数个数,其中0 ≤ x < 10n
利用组合的知识,假设一个整数有k位,从高到低每个位置如果不重复可以选取的数个数为9*9*8*(9 - k + 2),因为最高位不能为0,其他位均可。
代码:
int countNumbersWithUniqueDigits(int n) { if (n == 0) return 1; int len = n; int uniqueDigits[11]; uniqueDigits[1] = 9; for (int i = 1; i < 10; ++i) { uniqueDigits[i + 1] = uniqueDigits[i] * (9 - i + 1); } int res = 0; for (int i = 1; i <= n; ++i) { res += uniqueDigits[i]; } return res + 1; }
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