program to check if input string is valid anagram#260
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| import java.util.HashMap; | |||
| import java.util.Map; | |||
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Please name your class according to the filename.
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| import java.util.HashMap; | |||
| import java.util.Map; | |||
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Please name your file according to the class name.
| else if((hash1.size()==0 && hash2.size()==0) ) { | ||
| a = true; | ||
| } | ||
| else if(hash1.size()!=hash2.size()){ |
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unnecessary. You has check this condition above, too.
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| if(hash1.size()!=hash2.size()){ | ||
| a = false; |
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You can in this cases write return false;
| } | ||
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| else if((hash1.size()==0 && hash2.size()==0) ) { | ||
| a = true; |
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Same thing as above. return true;
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| for (int i = 0; i < s.length(); i++) { | ||
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| if (hash1.containsKey(s.charAt(i)) == false) { |
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Instead of == false use the not-operator !
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| for (int i = 0; i < t.length(); i++) { | ||
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| if (hash2.containsKey(t.charAt(i)) == false) { |
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Instead of == false use the not-operator !
| } | ||
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| else { | ||
| for (Map.Entry<Character, Integer> entry : hash1.entrySet()) { |
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Instead of this monster. You can use a = hash1.equals(hash2); For checking the equality of the HashMaps.
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| public static void main( String []args){ | ||
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| boolean b = isAnagram("a","ab"); |
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b is unnecessary. Because you can send the return value of the method right away to the println method.
| public class Solution { | ||
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| public static boolean isAnagram(String s, String t) { | ||
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Remove unnecessary empty lines.
| HashMap<Character, Integer> hash2 = new HashMap<>(); | ||
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| boolean a = false; | ||
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Put in some comments in your code.
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