357 Count Numbers with Unique Digits

1. Question

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])

2. Implementation

(1) Backtracking
class Solution {
public int countNumbersWithUniqueDigits(int n) {
boolean[] used = new boolean[10];
int[] count = new int[1];
getNumbersWithUniqueDigits(0, n, used, count);
// 加1因为要算上刚开始的0
return 1 + count[0];
}
public void getNumbersWithUniqueDigits(int curLevel, int depth, boolean[] used, int[] count) {
if (curLevel == depth) {
return;
}
int start = curLevel == 0 ? 1 : 0;
for (int i = start; i < 10; i++) {
if (!used[i]) {
used[i] = true;
++count[0];
getNumbersWithUniqueDigits(curLevel + 1, depth, used, count);
used[i] = false;
}
}
}
}
(2) DP
class Solution {
public int countNumbersWithUniqueDigits(int n) {
if (n > 10) {
return 0;
}
int[] dp = new int[11];
// 0个digit的时候,unique number是1, 这里我们把0当做
dp[0] = 1;
dp[1] = 9;
for (int i = 2; i <= n; i++) {
dp[i] = dp[i - 1] * (9 - i + 2);
}
int res = 0;
for (int i = 0; i <= n; i++) {
res += dp[i];
}
return res;
}
}

3. Time & Space Complexity

Backtracking: 时间复杂度O(n!), 如果n > 10的话,时间复杂度为O(10!), 空间复杂度O(n)
DP: 时间复杂度O(n), 空间复杂度O(1)