1178. Number of Valid Words for Each Puzzle

Posted 2022-09-20 lyhcc

题目说明

　　本题目是2019年9月1日LeetCode周赛的最后一题

题目描述

With respect to a given puzzle string, a word is valid if both the following conditions are satisfied: word contains the first letter of puzzle. For each letter in word, that letter is in puzzle 　　　　For example, if the puzzle is "abcdefg", then valid words are "faced", "cabbage", and "baggage"; while invalid words are "beefed" (doesn‘t include "a") and "based" (includes "s" which isn‘t in the puzzle). Return an array answer, where answer[i] is the number of words in the given word list words that are valid with respect to the puzzle puzzles[i].

样例

Input: 
words = ["aaaa","asas","able","ability","actt","actor","access"], 
puzzles = ["aboveyz","abrodyz","abslute","absoryz","actresz","gaswxyz"]
Output: [1,1,3,2,4,0]
Explanation:
1 valid word for "aboveyz" : "aaaa" 
1 valid word for "abrodyz" : "aaaa"
3 valid words for "abslute" : "aaaa", "asas", "able"
2 valid words for "absoryz" : "aaaa", "asas"
4 valid words for "actresz" : "aaaa", "asas", "actt", "access"
There‘re no valid words for "gaswxyz" cause none of the words in the list contains letter ‘g‘.

条件限制

1 <= words.length <= 10^5 4 <= words[i].length <= 50 1 <= puzzles.length <= 10^4 puzzles[i].length == 7 words[i][j], puzzles[i][j] are English lowercase letters. Each puzzles[i] doesn‘t contain repeated characters.

解题思路

① 先将words遍历一遍，然后将里面的每个单词的字母作为一个索引（即单词里面的字母-‘a’作为索引），将单词放到对应数组中 ② 计算单词对应的一个唯一的数值 　　我使用26个二进制表示，例如aba表示为11 　　1<<((对应不重复字母)-‘a’)计算一个字符对应的位置 　　而11 的第一个表示b 第二个表示为a ,重复字母不考虑 ③ 遍历puzzles数组，也计算相应的单词对应唯一的值，然后利用第一个字母查找符合题意的word（题意要求是，将puzzle里面的单词的首个字母包含于word中），然后遍历对应索引的word数组 ④ 判断word是否满足题目条件 　　而使用的方法是位运算中的与运算通过判断 puzzle对应的唯一值与word的唯一值进行与运算判断是否等于word的唯一值，满足则是满足条件 　　　　如 　　word ="a"  , puzzle  = "ab" 　　　　唯一值　　1　　　　　　11 　　　　&运算过后 为  1 　　这里需要注意的是：&运算的优先级低于==

代码

    vector<int>w;
    vector<int>wf[26];
    bool vis[26];
    vector<int> findNumOfValidWords(vector<string>& words, vector<string>& puzzles) 
        
        for(int i=0;i<words.size();i++)
            memset(vis,false,sizeof(vis));
            int tmp = 0;
            
            for(int j=0;j<words[i].size();j++)
                if(!vis[words[i][j]-‘a‘])
                    tmp+=(1<<(words[i][j]-‘a‘));
                    vis[words[i][j]-‘a‘]=true;
                    wf[words[i][j]-‘a‘].push_back(i);
                    
                
            

            w.push_back(tmp);
        
        
        vector<int>res;
        for(int i=0;i<puzzles.size();i++)
            int index = puzzles[i][0]-‘a‘;
            int h  = 0;
            for(int j=0;j<7;j++)
                h+=(1<<(puzzles[i][j]-‘a‘));
            
            int cnt =0 ;
            for(int j=0;j<wf[index].size();j++)        
                if((w[wf[index][j]]&h)==w[wf[index][j]])
                    cnt++;
                
            
            res.push_back(cnt);
        
        return res;
        
    

Code