标签：ued com bin because git i++ ati backtrack ber
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])
Analysis:
A number of unique digits is a number which is a combination of unrepeated digits. So, we can calculate the total number. for number with n digits, like 100-999 or 1000-9999, the total numbers with unique digits equals to 9*9*8...*(11-n). because the highest digit cannot be 0.
Here is DP. dp[i] is the count of all i-digit numbers with unique digits, dp[i] = dp[i-1]*(11-i) for i from 2 to n
1 public static int countNumbersWithUniqueDigits(int n) { 2 if (n == 0) { 3 return 1; 4 } 5 int ans = 10, base = 9; 6 for (int i = 2; i <= n && i <= 10; i++) { 7 base = base * (11 - i); 8 ans += base; 9 } 10 return ans; 11 }
第一遍backtracking做法：
1 public class Solution { 2 public int countNumbersWithUniqueDigits(int n) { 3 if (n == 0) return 1; 4 if (n == 1) return 10; 5 int res = 10; 6 for (int i=2; i<=n && i<=10; i++) { 7 int count = 1; 8 int bit = 9; 9 for (int j=0; j<i; j++) { 10 count *= bit; 11 if (j != 0) bit--; 12 } 13 res += count; 14 } 15 return res; 16 } 17 }
Leetcode: Count Numbers with Unique Digits
标签：ued com bin because git i++ ati backtrack ber
原文地址：http://www.cnblogs.com/EdwardLiu/p/6103510.html
