【Leetcode】Count Numbers with Unique Digits

最新推荐文章于 2022-04-11 15:44:00 发布

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最新推荐文章于 2022-04-11 15:44:00 发布

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分类专栏： leetcode Leetcode题解java版文章标签： leetcode

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本文介绍了一个LeetCode上的问题，即如何计算所有具有唯一数字的数的数量，这些数的范围为0到10^n-1。提供了两种解决方案：回溯法和动态规划方法，并详细解释了动态规划方法的实现。

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

题目：
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].

思路：
递推方程都给了，傻子采用回溯。。= =。。而且回溯看起来好难的样子。。

算法：

    public int countNumbersWithUniqueDigits(int n) {
        if (n == 0)
            return 1;
        int count = 0;
        for (int i = 1; i <= n; i++) {// 求每一位有多少unique digits
            count += cal(i);
        }
        return count;
    }

    public int cal(int k) {
        int sum = 9;
        if (k == 0)
            return 0;
        if (k == 1)
            return 10;
        for (int i = 11 - k; i <= 9; i++) {
            sum *= i;
        }
        return sum;
    }