leetcode 1079. Letter Tile Possibilities

Posted 2021-03-16 琴影

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。

 1 class Solution {
 2 private:
 3     void combination(const string &tiles, int len, vector<bool> &visited, string &temp, set<string> &st) {
 4         if (temp.length() > 0 && temp.length() <= len && st.find(temp) == st.end()) {
 5             st.insert(temp);
 6         }
 7         if (temp.length() > len)
 8             return;
 9         for (int i = 0; i < len; ++i) {
10             if (!visited[i]) {
11                 visited[i] = true;
12                 temp.push_back(tiles[i]);
13                 combination(tiles, len, visited, temp, st);
14                 visited[i] = false;
15                 temp.pop_back();
16             }
17         }
18     }
19 public:
20     int numTilePossibilities(string tiles) {
21         set<string> st;
22         int length = tiles.length();
23         string temp = "";
24         vector<bool> visited(length, false);
25         combination(tiles, length, visited, temp, st);
26         return st.size();
27     }
28 };

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