LeetCode：357. Count Numbers with Unique Digits

LeetCode：357. Count Numbers with Unique Digits

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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:

A direct way is to use the backtracking approach.
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.
This problem can also be solved using a dynamic programming approach and some knowledge of combinatorics.
Let f(k) = count of numbers with unique digits with length equals k.
f(1) = 10, …, f(k) = 9 * 9 * 8 * … (9 - k + 2) [The first factor is 9 because a number cannot start with 0].

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Hint写的很详细，找规律应该蛮好写。
注释掉的用找规律写的，后来看别人代码写了个回溯。其中用掩码来标识哪个数字用过，蛮有意思的。

Reference：回溯写法
http://www.cnblogs.com/grandyang/p/5582633.html

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class Solution {public:    int DFS(int pre, int max, int used) {        int res=0;       if(pre<max)  ++res;       else return res;       for(int i=0;i<10;++i)       {           if(!(used&(1<<i)))//判断是否使用过           {               used|=(1<<i);               int cur=10*pre+i;//竟然不能用pre=10*pre+i               res+=DFS(cur,max,used);               used&=~(1<<i);           }       }       return res;    }    //i=0表示0用过，=1，表示1用过。    int countNumbersWithUniqueDigits(int n) {        if(!n)  return 1;        int res = 1, max = pow(10, n), used = 0;        for(int i=1;i<10;++i)        {            used|=(1<<i);            res+=DFS(i,max,used);            used&=~(1<<i);        }        return res;        //if(n==1)    return 10;        // int t=9;        // for(int i=2;i<=n;++i)        // {        //     t*=(9-i+2);        // }        // return t+countNumbersWithUniqueDigits(n-1);    }};