leecode 357

最新推荐文章于 2025-09-04 12:26:54 发布

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最新推荐文章于 2025-09-04 12:26:54 发布

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文章标签： leetcode

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357. Count Numbers with Unique Digit

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])

几点注意：

1.数字为0—9，10个

2.n>10 是取n=10

3.rel[0] = 1;

rel[1] = 10;

rel[2] = rel[1] + (1~9)9 * (0~9去掉前面的)9

rel[3] = rel[2] + (1~9, 设取a)9*(0~9除了a，设取b)9*(0~9除ab)8

以此类推~~~~

class Solution {

public:
int countNumbersWithUniqueDigits(int n) {
if (n == 0)
return 1;
if(n == 1)
return 10;
if (n == 2)
return 91;
if (n > 10)
n = 10;
int* rel = new int [n];
int del = 9;
rel[0] = 1;
rel[1] = 10;
rel[2] = 10 + 9*9;
for (int i = 3 ; i <= n ; i ++)
{
rel[i] = 9;
for (int j = 1 ; j < i ; j ++)
rel[i] *= 10-j;
rel[i] += rel[i-1];
}
return rel[n];
}
};