leetcode 357. Count Numbers with Unique Digits

最新推荐文章于 2020-07-23 17:06:45 发布

原创最新推荐文章于 2020-07-23 17:06:45 发布 · 153 阅读

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本文介绍了一种计算特定范围内所有具有唯一数字组合的整数数量的方法。通过递减可用数字的数量来实现，从10个可能的数字开始，每次减少一个选项直到0，解决了重复数字的问题。文章还提供了一个具体的Java代码示例。

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])

数字排列组合的问题

public class Solution {
    public int countNumbersWithUniqueDigits(int n) {
        if(n==0) return 1;
        
        int ans = 10;
        int availableNum = 9;
        int uniqueCnt = 9;
        while(--n>0 && availableNum>0){
            uniqueCnt *= availableNum;
            ans += uniqueCnt;
            availableNum--;
        }
        return ans;
    }
}

贴上别人的解释

Following the hint. Let f(n) = count of number with unique digits of length n.

f(1) = 10. (0, 1, 2, 3, ...., 9)

f(2) = 9 * 9. Because for each number i from 1, ..., 9, we can pick j to form a 2-digit number ij and there are 9 numbers that are different from i for j to choose from.

f(3) = f(2) * 8 = 9 * 9 * 8. Because for each number with unique digits of length 2, say ij, we can pick k to form a 3 digit number ijk and there are 8 numbers that are different from i and j for k to choose from.

Similarly f(4) = f(3) * 7 = 9 * 9 * 8 * 7....

...

f(10) = 9 * 9 * 8 * 7 * 6 * ... * 1

f(11) = 0 = f(12) = f(13)....

any number with length > 10 couldn't be unique digits number.

The problem is asking for numbers from 0 to 10^n. Hence return f(1) + f(2) + .. + f(n)

因为11位数之后不会再有新的 unique 数字增加，可以建立一个数组用于查询