Code random notes_ Hash_ 1002 find common characters

Code Capriccio record two brush notes

LC1002. Find common characters

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subject

Here's an array of strings words , Please find out all in words A common character that appears in each string of （ Include repeating characters ）, And return as an array . You can press In any order Return to the answer .

Example 1：

Input ：words = [“bella”,“label”,“roller”]
Output ：[“e”,“l”,“l”]

Example 2：

Input ：words = [“cool”,“lock”,“cook”]
Output ：[“c”,“o”]

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Thought analysis

1. Topic meaning analysis ： from 26 Of the lowercase characters , Have a character (a) In all strings (aapply/aapple/aaugment) If it appears in , It outputs , Repeat it , The output [‘a’,‘a’],b,c,d,…z Empathy
2. Ideas ： Hash array , Record the hash value of each string , Minimum value
3. Deductive analysis ： Hash table statistics words in 26 Frequency of characters , Then take the minimum value of each string , Then convert to characters (String) Output

           a b     e             l     o     r               z
bella:    [1 1 0 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ]
label:    [1 1 0 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ]
roller:   [0 0 0 0 1 0 0 0 0 0 0 2 0 0 1 0 0 2 0 0 0 0 0 0 0 0 ]
Math.min: [0 0 0 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ]

 Get the output ["e","l","l"]

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Code implementation

Complete code implementation

public List<String> commonChars(String[] words) {
    
        List<String> res = new ArrayList<>();
        if(words.length == 0) return res;
        // Scan first String, initialization  hash
        int[] record = new int[26]; //  Record No.  1  individual String Of  Hash
        for (int i = 0; i < words[0].length();i++){
    
            record[words[0].charAt(i) - 'a']++;
        }
        // After scanning  String
        for (int i = 1; i < words.length;i++){
    
            int[] temp = new int[26]; //  Record No.  i  individual  String  Of  Hash
            for (int j = 0; j < words[i].length();j++){
    
                temp[words[i].charAt(j) - 'a']++;
            }
            // contrast   The first  i  individual  String  And the  i-1  individual  String  Of  Hash
            for(int k = 0; k < record.length;k++){
    
                record[k] = Math.min(record[k],temp[k]); take  Hash  minimum value 
            }
        }
        // Convert to  String
        for (int i = 0;i < record.length;i++) {
    
            while (record[i] > 0){
    
                // Because there is repetition , So we need to  while
                char c = (char)(i+'a');
                res.add(String.valueOf(c));
                record[i]--;
            }

        }
        return res;
    }

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