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])

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

分析

给定n, 给出0≤x<10n 之间没有重复位数的数字的个数.

用排列组合的规律可以求出来:
10 + 9 * 9 + 9 * 9 * 8 + 9 * 9 * 8 * 7 + … (加到第10项之后就不必再加了.)

实现

0ms

class Solution {public:    int countNumbersWithUniqueDigits(int n) {        if (n == 0) return 1;        int result = 10, pdt = 9;        for (int i = 1; i < n; i++) {            pdt *= (10 - i);            result += pdt;            if (i > 10) break; // 位数大于10必定有重复.不存在满足条件的数        }        return result;    }};