[LeetCode] Count Numbers with Unique Digits 计算各位不相同的数字个数

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].

Credits:
Special thanks to @memoryless for adding this problem and creating all test cases.

class Solution {
public:
int countNumbersWithUniqueDigits(int n) {
if (n == 0) return 1;
int res = 0;
for (int i = 1; i <= n; ++i) {
res += count(i);
}
return res;
}
int count(int k) {
if (k < 1) return 0;
if (k == 1) return 10;
int res = 1;
for (int i = 9; i >= (11 - k); --i) {
res *= i;
}
return res * 9;
}
};

class Solution {
public:
int countNumbersWithUniqueDigits(int n) {
if (n == 0) return 1;
int res = 10, cnt = 9;
for (int i = 2; i <= n; ++i) {
cnt *= (11 - i);
res += cnt;
}
return res;
}
};

class Solution {
public:
int countNumbersWithUniqueDigits(int n) {
int res = 1, max = pow(10, n), used = 0;
for (int i = 1; i < 10; ++i) {
used |= (1 << i);
res += search(i, max, used);
used &= ~(1 << i);
}
return res;
}
int search(int pre, int max, int used) {
int res = 0;
if (pre < max) ++res;
else return res;
for (int i = 0; i < 10; ++i) {
if (!(used & (1 << i))) {
used |= (1 << i);
int cur = 10 * pre + i;
res += search(cur, max, used);
used &= ~(1 << i);
}
}
return res;
}
};

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