1079. 活字印刷

你有一套活字字模 tiles，其中每个字模上都刻有一个字母 tiles[i]。返回你可以印出的非空字母序列的数目。

注意：本题中，每个活字字模只能使用一次。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/letter-tile-possibilities
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

回溯

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;

class Solution {

    private int solve(Map<Character, Integer> cntMap, Set<Character> characterSet, int n) {
        if (n == 0) {
            return 1;
        }
        int result = 1;
        for (Character character : characterSet) {
            if (cntMap.get(character) > 0) {
                cntMap.merge(character, -1, Integer::sum);
                result += solve(cntMap, characterSet, n - 1);
                cntMap.merge(character, 1, Integer::sum);
            }
        }
        return result;
    }

    public int numTilePossibilities(String tiles) {
        if (tiles == null || tiles.length() == 0) {
            return 0;
        }
        Map<Character, Integer> cntMap = new HashMap<>();
        for (int i = 0; i < tiles.length(); i++) {
            cntMap.merge(tiles.charAt(i), 1, Integer::sum);
        }
        Set<Character> characterSet = cntMap.keySet();
        return solve(cntMap, characterSet, tiles.length()) - 1;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNext()) {
            System.out.println(new Solution().numTilePossibilities(in.next()));
        }
    }
}

DP

class Solution {

    private static final int[][] c = new int[8][8];

    static {
        for (int i = 0; i < 8; i++) {
            c[i][0] = c[i][i] = 1;
            for (int j = 1; j < i; ++j) {
                c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
            }
        }
    }

    public int numTilePossibilities(String tiles) {
        if (tiles == null || tiles.length() == 0) {
            return 0;
        }
        Map<Character, Integer> cntMap = new HashMap<>();
        for (int i = 0; i < tiles.length(); i++) {
            cntMap.merge(tiles.charAt(i), 1, Integer::sum);
        }
        int m = cntMap.size();
        int n = tiles.length();

        int[][] f = new int[m + 1][n + 1];
        f[0][0] = 1;
        int i = 0;
        for (int cnt : cntMap.values()) {
            i++;
            for (int j = 0; j <= n; j++) {
                for (int k = 0; k <= cnt && k <= j; k++) {
                    f[i][j] += f[i - 1][j - k] * c[j][k];
                }
            }
        }
        return Arrays.stream(f[m]).sum() - f[m][0];
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNext()) {
            System.out.println(new Solution().numTilePossibilities(in.next()));
        }
    }
}