Java [Leetcode 357]Count Numbers with Unique Digits

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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])

解题思路：

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

代码如下：

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

Java [Leetcode 357]Count Numbers with Unique Digits

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原文地址：http://www.cnblogs.com/zihaowang/p/5731238.html