leetcode_Count Numbers with Unique Digits

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sunp823

最新推荐文章于 2025-04-21 22:09:55 发布

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本文介绍了一种通过迭代方法计算特定范围内具有唯一数字组合的数量的方法。对于非负整数n，计算所有数字x的总数，使得0≤x<10^n且x的所有位数都不相同。文中提供了一个C++实现示例。

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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])
Hint:
1.A direct way is to use the backtracking approach.
2.Backtracking should contains three states which are (the current number, number of steps to get that number and a bitmask which represent which number is marked as visited so far in the current number). Start with state (0,0,0) and count all valid number till we reach number of steps equals to 10n.
3.This problem can also be solved using a dynamic programming approach and some knowledge of combinatorics.
4.Let f(k) = count of numbers with unique digits with length equals k.
5.f(1) = 10, …, f(k) = 9 * 9 * 8 * … (9 - k + 2) [The first factor is 9 because a number cannot start with 0].
1.[解法]：
题目的hint已经说得很清楚了，可以采用一种迭代的方式解决，f(k)代表k位数（0<=N<10^k）内各个位上具有不同数字的数字的个数，则具有以下迭代关系：
f(1) = 10;
f(k) = 9 * 9 * 8 * … (9 - k + 2);注意第一个9是固定存在的。
最后我们求得的result = f(1) + f(2) + … +f(n);
2.【代码】如下：

#include<iostream>
using namespace std;
int countNumbersWithUniqueDigits(int n)
{
    int result = 0;
    if(n == 0)
    return 1;
    int *arr = new int[n+1];
    arr[1] = 10;
    for(int i = 2;i<=n;i++)
    {
        int temp = 9;
        for(int j = 2;j <= i;j++)
        {
            temp*=(9 - j + 2);
        }
        arr[i] = temp; 
    }
    for(int i = 1;i<=n;i++)
    {
        result += arr[i];
    }
    delete arr;
    return result;       
}
int main(void)
{

    cout<<countNumbersWithUniqueDigits(0)<<endl;
    cout<<countNumbersWithUniqueDigits(1)<<endl;
    cout<<countNumbersWithUniqueDigits(2)<<endl;
    system("pause");
    return 0;
}