Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.
Example:
Input: 2
Output: 91
Explanation: The answer should be the total numbers in the range of 0 ≤ x < 100,
excluding 11, 22, 33, 44, 55, 66, 77, 88, 99
Approach #1: Math. [C++]
class Solution {
public:
int countNumbersWithUniqueDigits(int n) {
if (n == 0) return 1;
if (n == 1) return 10;
int ans = 10;
int count = 9;
int temp = 9;
while (n > 1 && count > 0) {
temp = temp * count;
ans += temp;
count--;
n--;
}
return ans;
}
};
Analysis:
f(1) = 10;
f(2) = f(1) + 9 * 9; xy: x: can be 1, 2.....8, 9. y: can‘t be the same as with x but it can be 0, so there are 9 difference kinds choise.
f(2) = f(2) + 9 * 9 * 8;
Approach #2: backtracking. [C++]
public class Solution {
public static int countNumbersWithUniqueDigits(int n) {
if (n > 10) {
return countNumbersWithUniqueDigits(10);
}
int count = 1; // x == 0
long max = (long) Math.pow(10, n);
boolean[] used = new boolean[10];
for (int i = 1; i < 10; i++) {
used[i] = true;
count += search(i, max, used);
used[i] = false;
}
return count;
}
private static int search(long prev, long max, boolean[] used) {
int count = 0;
if (prev < max) {
count += 1;
} else {
return count;
}
for (int i = 0; i < 10; i++) {
if (!used[i]) {
used[i] = true;
long cur = 10 * prev + i;
count += search(cur, max, used);
used[i] = false;
}
}
return count;
}
}
357. Count Numbers with Unique Digits
原文：https://www.cnblogs.com/ruruozhenhao/p/10353330.html