LC 357. Count Numbers with Unique Digits

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

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Count Numbers with Unique Digits.

额有点蠢。

f(k) = 9 * 9 * 8 * ... * (9 - i + 2) 第一位是9 因为 0 不能在第一位。

class Solution {
public:
    int countNumbersWithUniqueDigits(int n) {
      if(n == ) return ;
      vector<int> dp(n+, );
      dp[] = ;
      for(int i=; i<n+; i++){
        dp[i] = dp[i-] * (-i+);
      }
      int ret = ;
      for(int i=; i<n+; i++) ret += dp[i];
      ret += ;
      return ret;
    }

};

DFS

Runtime: 116 ms, faster than 2.80% of C++ online submissions for Count Numbers with Unique Digits.

class Solution {
public:
    int countNumbersWithUniqueDigits(int n) {
      int maxval = pow(,n), ret = , used = ;
      for(int i=; i<; i++){
        used |= (<<i);
        search(i, maxval, ret, used);
        used &= ~(<<i);
      }
      return ret;
    }
      void search(int prev, int maxval, int& ret, int& used){
        if(prev < maxval) ret++;
        else return ;
        for(int i=; i<; i++){
          if(!(used & ( << i))){
            used |= (<<i);
            search(*prev+i, maxval, ret, used);
            used &= ~(<<i);
          }
        }
      }
};