README_EN.md

357. Count Numbers with Unique Digits

中文文档

Description

Given an integer n, return the count of all numbers with unique digits, x, where 0 <= x < 10n.

 

Example 1:

Input: n = 2
Output: 91
Explanation: The answer should be the total numbers in the range of 0 ≤ x < 100, excluding 11,22,33,44,55,66,77,88,99

Example 2:

Input: n = 0
Output: 1

 

Constraints:

• 0 <= n <= 8

Solutions

Python3

class Solution:
    def countNumbersWithUniqueDigits(self, n: int) -> int:
        if n == 0:
            return 1
        if n == 1:
            return 10
        ans, cur = 10, 9
        for i in range(n - 1):
            cur *= (9 - i)
            ans += cur
        return ans

Java

class Solution {
    public int countNumbersWithUniqueDigits(int n) {
        if (n == 0) {
            return 1;
        }
        if (n == 1) {
            return 10;
        }
        int ans = 10;
        for (int i = 0, cur = 9; i < n - 1; ++i) {
            cur *= (9 - i);
            ans += cur;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countNumbersWithUniqueDigits(int n) {
        if (n == 0) return 1;
        if (n == 1) return 10;
        int ans = 10;
        for (int i = 0, cur = 9; i < n - 1; ++i)
        {
            cur *= (9 - i);
            ans += cur;
        }
        return ans;
    }
};

Go

func countNumbersWithUniqueDigits(n int) int {
	if n == 0 {
		return 1
	}
	if n == 1 {
		return 10
	}
	ans := 10
	for i, cur := 0, 9; i < n-1; i++ {
		cur *= (9 - i)
		ans += cur
	}
	return ans
}

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