357. Count Numbers with Unique Digits 类别：动态规划 难度：medium

题目：

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]

)

思路：

当n=1时因为只有一个数字，所以0-9都是答案．当n>=2时，最高位可以为1-9任意一个数字，之后各位可以选择的数字个数依次为9, 8, 7, 6...，上一位选一个下一位就少了一种选择．

程序：

class Solution {
public:
int countNumbersWithUniqueDigits(int n) {

if(n == 0)
return 1;
if(n <= 10)
{
vector<int> res(n,1);
for(int i = 1;i <= n;i++)
{
int sum = 9;
for(int j = 2;j <= i;j++)
sum *= (11 - j);

res[i] = res[i - 1] + sum;
}
return res
;
}
else
{
vector<int> res(10,1);
for(int i = 1;i <= 10;i++)
{
int sum = 9;
for(int j = 2;j <= i;j++)
sum *= (11 - j);

res[i] = res[i - 1] + sum;
}
return res[10];
}

}
};