LeetCode解题报告 357. Count Numbers with Unique Digits [medium]

题目描述

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].

解题思路

题目就是给定一个n，返回［0 ，10n）之间的所有各个位数不相同的数字的个数。
利用提示5的思路，结合排列组合的知识求解。
给定一个k位的数，要求各个位数不相同，从左至右每个位数能选择的数字依次为9（首位不能为0），9，8，7，6，5...
求所有k位数的这样的数的数目之和，即为从1位，2位，3位...一直加到n位的和。

如k＝1，0-9共10个数；
k＝2，k＝1的10个数＋9*9＝10+81=91；
k＝3，k＝1的10个数＋k＝2的91个数＋9*9*8＝10＋91+648＝739
...

注意，当n＝0时，返回1.

时间复杂度分析

从第n位循环到底n-(n-2)位，时间复杂度为O(N).

代码如下：

class Solution {
public:
int countNumbersWithUniqueDigits(int n) {
if (n==0) {
return 1;
}
if (n==1) {
return 10;
}
else{
int raise=0;
int result=10;
int temp=9;
raise=9;
for (int i=2; i<=n; i++) {
raise=raise*temp;
temp--;
result=result+raise;
}
return result;
}
}
};