leecode 357

357. Count Numbers with Unique Digit

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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])

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];
    }
};