357. Count Numbers with Unique Digits
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:
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A direct way is to use the backtracking approach.
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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.
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This problem can also be solved using a dynamic programming approach and some knowledge of combinatorics.
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Let f(k) = count of numbers with unique digits with length equals k.
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f(1) = 10, ..., f(k) = 9 * 9 * 8 * ... (9 - k + 2) [The first factor is 9 because a number cannot start with 0].
题目大意：
找出10的n次方内，没有重复数字的数的个数。例如10的3次方内，102为合法值，101为非法值。
思路：
采用排列组合来求出10的i次方，比如10的平方，范围为[1,100)，然后找出这个范围内合法值有几个。9*9(第一位不能为0，所以为9，第二位可以为除了第一位以外的9中情况)。
| n次方 | 范围 | 合法个数 |
| 0 | [0,1) | 1 |
| 1 | [1,10) |
9 |
| 2 | [10,100) | 9*9 |
| 3 |
[100,1000) | 9*9*8 |
| ... | ... | ... |
| i（i<9） |
[10的i-1次方，10的i次方) | 9*9*8*7*...*(9 - n + 2) |
| 9 |
[100000000,1000000000) | 9*9*8*7*6*5*4*3*2 |
经过上面分析，当n大于等于10的时候，合法值不再增加，因为n>=10时，数的位数超过了10位，所以肯定有重复的数字。
代码如下：
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class
Solution {
public
:
int
countNumbersWithUniqueDigits(
int
n) {
int
result,tmp;
if
(0 == n)
return
1;
if
(1 == n)
return
10;
result = 10;
tmp = 9;
for
(
int
i = 2; i<=min(n,9); ++i)
{
result += tmp * (11 - i);
tmp *= (11 - i);
}
return
result;
}
};
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