Leetcode: Count Numbers with Unique Digits

Posted 2020-08-21 neverlandly

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])

Analysis:

A number of unique digits is a number which is a combination of unrepeated digits. So, we can calculate the total number. for number with n digits, like 100-999 or 1000-9999, the total numbers with unique digits equals to 9*9*8...*(11-n). because the highest digit cannot be 0.

Here is DP. dp[i] is the count of all i-digit numbers with unique digits, dp[i] = dp[i-1]*(11-i) for i from 2 to n

 1 public static int countNumbersWithUniqueDigits(int n) {
 2     if (n == 0) {
 3         return 1;
 4     }
 5     int ans = 10, base = 9;
 6     for (int i = 2; i <= n && i <= 10; i++) {
 7         base = base * (11 - i);
 8         ans += base;
 9     }
10     return ans;
11 }

 

第一遍backtracking做法：

 1 public class Solution {
 2     public int countNumbersWithUniqueDigits(int n) {
 3         if (n == 0) return 1;
 4         if (n == 1) return 10;
 5         int res = 10;
 6         for (int i=2; i<=n && i<=10; i++) {
 7             int count = 1;
 8             int bit = 9;
 9             for (int j=0; j<i; j++) {
10                 count *= bit;
11                 if (j != 0) bit--;
12             }
13             res += count;
14         }
15         return res;
16     }
17 }