LeetCode刷题笔记 1079. 活字印刷（？）

题目描述

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。

示例：
输入：“AAB”
输出：8
解释：可能的序列为 “A”, “B”, “AA”, “AB”, “BA”, “AAB”, “ABA”, “BAA”。

总结

SC 回溯的题有些用数学是投机取巧，但有些不是，是6
SC2 就是抽象出来，不在乎实际排列出来是什么字符串，而是 计数 判断

Sample Code 1 (不理解)

class Solution {
    public int numTilePossibilities(String tiles) {
        char[] chars = tiles.toCharArray();
        Arrays.sort(chars);
        return helper(0, 1, 1, 1, new int[26], chars);
    }


    private int helper(int start, int a, int b, int m, int[] counts, char[] chars) {
        int result = 0;
        a *= m;
        
        for (int i = start; i < chars.length; i++) {
            int c_index = chars[i] - 'A';
            if (i == start || chars[i] != chars[i - 1]) {
                b *= (++counts[c_index]);
                result += a / b + helper(i + 1, a, b, m+1, counts, chars);
                b /= (counts[c_index]--);
            }
        }
        return result;
    }
}

Sample Code 2

class Solution {
    public int numTilePossibilities(String tiles) {
        int[] record = new int[26];
        for (char c : tiles.toCharArray()) {
            record[c - 'A']++;
        }
        return dfs(record);
    }

    private int dfs(int[] record) {
        int sum = 0;
        for (int i = 0; i < 26; i++) {
            if (record[i] == 0) continue;
            record[i]--;
            sum++;
            sum += dfs(record);
            record[i]++;
        }
        return sum;
    }
}

Demo Code

class Solution {
    List<String> list = new ArrayList<>();
    public int numTilePossibilities(String tiles) {
        boolean[] visited = new boolean[tiles.length()+1];
    
        find(new StringBuilder(), visited, tiles);
        return list.size();            
    }
    
    private void find(StringBuilder temp, boolean[] visited, String tiles) {   
        if(temp.length() >= tiles.length()) {
            return;
        }
        
        for(int i = 0; i < tiles.length(); i++) {
            if(visited[i] == false) {
                temp.append(tiles.substring(i, i+1));
                visited[i] = true;
                if(!list.contains(temp.toString())) {
                    list.add(temp.toString());
                    find(temp, visited, tiles);
                }
                temp.delete(temp.length()-1, temp.length());
                visited[i] = false;
            }
        }
    } 
}

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/letter-tile-possibilities