leetcode 1079. Letter Tile Possibilities

You have a set of tiles, where each tile has one letter tiles[i]printed on it.  Return the number of possible non-empty sequences of letters you can make.

 

Example 1:

Input: "AAB"
Output: 8
Explanation: The possible sequences are "A", "B", "AA", "AB", "BA", "AAB", "ABA", "BAA".

Example 2:

Input: "AAABBC"
Output: 188

 

Note:

1. 1 <= tiles.length <= 7
2. tiles consists of uppercase English letters.

 

思路一：根据给定字符串，生成所有的序列方式可以直接利用回溯法生成，但是由于有些字符重复，所以会有重复的子序列生成，因此我们可以用set处理。

class Solution {
private:
    void combination(const string &tiles, int len, vector<bool> &visited, string &temp, set<string> &st) {
        if (temp.length() > 0 && temp.length() <= len && st.find(temp) == st.end()) {
            st.insert(temp);
        }
        if (temp.length() > len)
            return;
        for (int i = 0; i < len; ++i) {
            if (!visited[i]) {
                visited[i] = true;
                temp.push_back(tiles[i]);
                combination(tiles, len, visited, temp, st);
                visited[i] = false;
                temp.pop_back();
            }
        }
    }
public:
    int numTilePossibilities(string tiles) {
        set<string> st;
        int length = tiles.length();
        string temp = "";
        vector<bool> visited(length, false);
        combination(tiles, length, visited, temp, st);
        return st.size();
    }
};

思路二：

class Solution {
private:
    int dfs(vector<int> &cnt) {
        int num = 0;
        for (int i = 0; i < 26; ++i) {
            if (cnt[i] == 0) continue;
            num++;
            cnt[i]--;
            num += dfs(cnt);
            cnt[i]++;
        }
        return num;
    }
public:
    int numTilePossibilities(string tiles) {
        vector<int> cnt(26, 0);
        for (auto c: tiles) {
            cnt[c - 'A']++;
        }
        return dfs(cnt);
    }
};

这种做法保证了在子序列字符串的某一位字符选择上不会重复选择