Code random notes_ Hash_ 242 effective letter heterotopic words

Code Capriccio record two brush notes

LC242. Effective alphabetic words

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subject

Given two strings s and t , Write a function to determine t Whether it is s Letter heteronym of .

Be careful ： if s and t Each character in the has the same number of occurrences , said s and t They are mutually alphabetic words .

Example 1:

Input : s = “anagram”, t = “nagaram”
Output : true

Example 2:

Input : s = “rat”, t = “car”
Output : false

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Thought analysis

because s and t Only lowercase letters ,26 The letters are fixed , You can build an array of records record, take s Record the number of times each character appears in ( With + In the form of ) , After then t Record the number of times each character appears in ( With - In the form of ) .

If it's an alphabetic word , Then the final record array must be all 0, Otherwise, it is not an alphabetic ectopic word .

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Code implementation

Complete code implementation

public boolean isAnagram(String s, String t) {
    
	// take  s  and  t  Convert to char Array  
    char[] sca = s.toCharArray();
    char[] tca = t.toCharArray();
    // Maintain a hash dictionary 
    int[] dict = new int[26];
    // take  s  Characters in  +  A formal record of 
    for(char c:sca){
    
        dict[c - 'a']++;
    }
    // take  t  Characters in  -  A formal record of 
    for(char c:tca){
    
        dict[c - 'a']--;
    }
    // If the final dictionary is  0  It represents the letter ectopic word required by the title 
    for(int i:dict){
    
        if(i != 0) return false;
    }
    return true;
}

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