【LEETCODE】模拟面试-357- Count Numbers with Unique Digits

题目:

https://leetcode.com/problems/count-numbers-with-unique-digits/

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].

分析:

This question is to get a count. Given n, count the numbers which belong to [0, 10^n) and do not contain duplicate digits.

Firstly, this n must be less or equal to 10, since there are 0~9 10 unique digits.

When n=1, count=10. (0~9) When n=2, there are 2 classes of numbers, one is 1-digit, another is 2-digit. Under this n, when i==1, count=10. when i==2, count=9*(8+1). when i==3, count=9*9*(7+1) ... when i==n, count=9*9*8..*(9-n+2)

So here we can use Dynamic Programming. Let dp[] to be an array with length=1*(n+1). dp[k] means if a number is of k digits, how many kinds of combinations can satisfy the requirement. For n, the number may have k=1~n situations. For k, the choices on the (i)th position depends on the (i-1)th and before. Finally, we will sum the dp from dp[0]~dp[n], and this is the result.

Python

class Solution(object):
    def countNumbersWithUniqueDigits(self, n):
        """
        :type n: int
        :rtype: int
        """
      
        n = min(n, 10)
        dp = [1] + [9]*n
        
        for k in xrange(2, n+1):
            for i in xrange(9, 9-k+1, -1):
                dp[k] *= i
        
        return sum(dp)

本文参与腾讯云自媒体分享计划,欢迎正在阅读的你也加入,一起分享。

发表于

我来说两句

0 条评论
登录 后参与评论

相关文章

来自专栏ml

HDUOJ--1159Common Subsequence

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536...

27570
来自专栏ml

HDUOJ----2647Reward

Reward Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Ja...

37780
来自专栏ml

HDUOJ--------A simple stone game(尼姆博弈扩展)(2008北京现场赛A题)

A simple stone game                                                             ...

29450
来自专栏ml

HDUOJ------Daydream字符查找-并求其始末位置

2013-07-17  10:50:38 Daydream Time Limit: 2000/1000 MS (Java/Others)    Memory L...

29460
来自专栏ml

HDUOJ---What Are You Talking About

What Are You Talking About Time Limit: 10000/5000 MS (Java/Others)    Memory Lim...

405140
来自专栏小樱的经验随笔

HDU 1004 Let the Balloon Rise【STL<map>】

Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 655...

31440
来自专栏算法修养

HDU-1054 Strategic Game(树形DP)

Strategic Game Time Limit : 20000/10000ms (Java/Other) Memory Limit : 65536/32...

35380
来自专栏ml

HDUOJ 1099——Lottery

Lottery Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (J...

37470
来自专栏算法修养

HDU 5651 xiaoxin juju needs help

xiaoxin juju needs help Time Limit: 2000/1000 MS (Java/Others)    Memory Limit:...

30570
来自专栏ml

hdu-----(1507)Uncle Tom's Inherited Land*(二分匹配)

Uncle Tom's Inherited Land* Time Limit: 2000/1000 MS (Java/Others)    Memory Lim...

26840

扫码关注云+社区

领取腾讯云代金券