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 eq...
Note that the digits'0'and'1'do not map to any letters, so Alicedoes notuse them. However, due to an error in transmission, Bob did not receive Alice's text message but received astring of pressed keysinstead. For example, when Alice sent the message"bob", Bob received the string"226...
Can you solve this real interview question? Count Numbers with Unique Digits - Given an integer n, return the count of all numbers with unique digits, x, where 0 <= x < 10n. Example 1: Input: n = 2 Output: 91 Explanation: The answer should be the t
代码: publicclassSolution {publicintcountNumbersWithUniqueDigits(intn) {if(0 ==n)return1;if(1 ==n)return10;intsum = 10;for(inti=2; i<=n; i++) sum+=getMul(i);returnsum; }//计算f(k)publicintgetMul(intn){intresult = 1;for(inti=9-n+2; i<=9; i++) result*=i;returnresu...
leetcode 357. Count Numbers with Unique Digits 统计独特编码数组的数量 + 一个很简单的排列组合问题 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 ...
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]. Credits: Special thanks to@memorylessfor adding this problem and creating all test...
Count the number of prime numbers less than a non-negative number, n. Example: Input: 10 Output: 4 Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7. 题目大意 # 统计所有小于非负整数 n 的质数的数量。 解题思路 # 给出一个数字 n,要求输出小于 n 的所有素数...
packageleetcode// 暴力打表法funccountNumbersWithUniqueDigits1(nint)int{res:=[]int{1,10,91,739,5275,32491,168571,712891,2345851,5611771,8877691}ifn>=10{returnres[10]}returnres[n]}// 打表方法funccountNumbersWithUniqueDigits(nint)int{ifn==0{return1}res,uniqueDigits,availableNumber:=10,9,9forn...
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]. 题目大意: 找出10的n次方内,没有重复数字的数的个数。例如10的3次方内,102为合法...
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