[LeetCode] 1079. Letter Tile Possibilities

Posted 2023-05-19 CNoodle

You have n  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 using the letters printed on those tiles.

Example 1:

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

Example 2:

Input: tiles = "AAABBC"
Output: 188

Example 3:

Input: tiles = "V"
Output: 1 

Constraints:

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

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。

思路是 DFS backtracking。这道题有点类似求子集那一类的题。因为 input 中有可能有重复字母，为了不重复使用，我们需要将 input 字符串中的字母进行排序，再通过一个额外的 visited 数组判断某个字母是否有被重复使用。

时间O(nlogn)

空间O(n)

Java实现

 class Solution 
     int count;
 
     public int numTilePossibilities(String tiles) 
         count = 0;
         char[] letters = tiles.toCharArray();
         Arrays.sort(letters);
         boolean[] visited = new boolean[letters.length];
         dfs(letters, 0, visited);
         return count;
     
 
     private void dfs(char[] letters, int index, boolean[] visited) 
         if (index == letters.length) 
             return;
         
         for (int i = 0; i < letters.length; i++) 
             if (visited[i]) 
                 continue;
             
             if (i - 1 >= 0 && letters[i] == letters[i - 1] && !visited[i - 1]) 
                 continue;
             
             count++;
             visited[i] = true;
             dfs(letters, index + 1, visited);
             visited[i] = false;
         
     
 

 

LeetCode 题目总结

LeetCode（17）Letter Combinations of a Phone Number

题目如下：

Python代码：

class Solution(object):
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        if not digits:
            return []
        dic = {\'2\':\'abc\',\'3\':\'def\',\'4\':\'ghi\',\'5\':\'jkl\',\'6\':\'mno\',\'7\':\'pqrs\',\'8\':\'tuv\',\'9\':\'wxyz\'}
        result = []
        self.helper(digits,dic,0,"",result)
        return result
    
    def helper(self,digits,dic,index,temp,result):
        if index==len(digits):
            result.append(temp)
        else:
            s = dic[digits[index]]
            for i in s:
                temp += i
                self.helper(digits,dic,index+1,temp,result)
                temp = temp[:-1]