Leetcode 357. Count Numbers with Unique Digits

357. Count Numbers with Unique Digits

• Total Accepted: 10844
• Total Submissions: 25103
• Difficulty: Medium

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].

思路：Hint中提示的一样，既可以用回溯，也可以用DP的递推。

代码：

方法一：

n=0:  f(0)=1

n=1:  f(1)=10

2<=n<=10:  f(n)=9*9*8*...*(9-k+2)+f(k-1)

n>10:  f(n)=f(10)

class Solution {
public:
    int countNumbersWithUniqueDigits(int n) {
        if(!n) return 1;
        int res=10,mutil=9;
        n=n>10?10:n;
        for(int i=1;i<n;i++){
            mutil*=10-i;
            res+=mutil;
        }
        return res;
    }
};

方法二：回溯法。

java版的：

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;
    }
}

不过下面的c++版，运行超时。

class Solution {
public:
    int countNumbersWithUniqueDigits(long long cur,long long maxval,vector<bool> exist){
        if(cur>=maxval) return 0;
        int count=1;
        for(int i=0;i<10;i++){
            if(!exist[i]){
                exist[i]=true;
                count+=countNumbersWithUniqueDigits(cur*10+i,maxval,exist);
                exist[i]=false;
            }
        }
        return count;
    }
    int countNumbersWithUniqueDigits(int n){
        int count=1;
        long long maxval=pow(10,n);
        n=n>10?10:n;
        vector<bool> exist(10,false);
        for(int i=1;i<10;i++){
            exist[i]=true;
            count+=countNumbersWithUniqueDigits(i,maxval,exist);           
            exist[i]=false;
        }
        return count;
    }
};