357. Count Numbers with Unique Digits

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本文介绍了一种通过动态规划和组合数学原理来计算特定范围内所有具有唯一数字的整数数量的方法。通过递归公式计算不同长度下唯一数字的整数数量，并提供了具体的Java实现。

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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:

A direct way is to use the backtracking approach.
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.
This problem can also be solved using a dynamic programming approach and some knowledge of combinatorics.
Let f(k) = count of numbers with unique digits with length equals k.
f(1) = 10, ..., f(k) = 9 * 9 * 8 * ... (9 - k + 2) [The first factor is 9 because a number cannot start with 0].

提示已经给出，找规律数有点费劲还容易晕，直接看做组合问题就方便了。

需要取每一位都不同的数，一位数可取0~9，第二位数还剩下9个，第三位数还剩下8个……以此类推。

当然组成的时候比如三位数，是不能有前导0的，所以把0抓出来单独处理。两位的时候9 *9 ，三位的时候9*9*8……

public static int countNumbersWithUniqueDigits(int n)
	{
		if(n==0)
			return 1;
		if(n==1)
			return 10;
		
		int[] arr=new int[n+1];
		arr[0]=1;
		arr[1]=10;
		
		for(int i=2;i<=n;i++)
		{
			int factor=9;
			for(int k=9;k>=9-i+2;k--)
				factor=factor*k;
			arr[i]=factor;
		}
		
		int sum=0;
		for(int i=1;i<=n;i++)
			sum+=arr[i];
			
		return sum;
	}