Numbers At Most N Given Digit Set (H)

题目

Given an array of digits, you can write numbers using each digits[i] as many times as we want. For example, if digits = ['1','3','5'], we may write numbers such as '13', '551', and '1351315'.

Return the number of positive integers that can be generated that are less than or equal to a given integer n.

Example 1:

Input: digits = ["1","3","5","7"], n = 100
Output: 20
Explanation: 
The 20 numbers that can be written are:
1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.

Example 2:

Input: digits = ["1","4","9"], n = 1000000000
Output: 29523
Explanation: 
We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers,
81 four digit numbers, 243 five digit numbers, 729 six digit numbers,
2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers.
In total, this is 29523 integers that can be written using the digits array.

Example 3:

Input: digits = ["7"], n = 8
Output: 1

Constraints:

• 1 <= digits.length <= 9
• digits[i].length == 1
• digits[i] is a digit from '1' to '9'.
• All the values in digits are unique.
• 1 <= n <= 109

题意

用给定的数字组成一个正整数，使得它比目标值小，统计这样的正整数的个数。

思路

记目标值的位数为len，如果组成的正整数k长度小于len，那么可以随意组合；如果长度等于len，那么对第一位进行比较，如果k的第一位小于目标值的第一位，那么剩余位可以随意组合，如果第一位相同，递归求解即可。

代码实现

Java

class Solution {
    public int atMostNGivenDigitSet(String[] digits, int n) {
        Arrays.sort(digits);
        return helper(digits, n, false);
    }

    private int helper(String[] digits, int n, boolean fillAll) {
        if (n == 0) {
            return 0;
        }

        int count = 0;
        String ns = n + "";
        int len = ns.length();
        int size = digits.length;

      	// 仅在第一次调用该方法时需要累加
        if (!fillAll) {
            for (int i = 1; i < len; i++) {
                count += (int) Math.pow(size, i);
            }
        }

        for (int i = 0; i < size; i++) {
            if (digits[i].charAt(0) < ns.charAt(0)) {
                count += (int) Math.pow(size, len - 1);
            } else if (digits[i].charAt(0) == ns.charAt(0) && len == 1) {
                count += 1;
                break;
            } else if (digits[i].charAt(0) == ns.charAt(0) && ns.charAt(1) != '0') {
                count += helper(digits, Integer.parseInt(ns.substring(1)), true);
                break;
            }
        }

        return count;
    }
}