【Leetcode】Count Numbers with Unique Digits

发布时间：2020-12-13 21:11:38 所属栏目：PHP教程来源：网络整理

导读：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 number

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) 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，傻子采取回溯。。= =。。而且回溯看起来好难的模样。。

算法：

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

（编辑：李大同）

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