"Leetcode-Interview algorithm classic-java Implementation" "205-isomorphic Strings (isomorphic string)"

"205-isomorphic Strings (Homogeneous string)" "leetcode-Interview algorithm classic-java Implementation" "All topics Directory Index" code Download "Https://github.com/Wang-Jun-Chao" Original Question

Given strings S and T, determine if they are isomorphic.
The strings is isomorphic if the characters in S can is replaced to get t.
All occurrences of a character must is replaced with another character while preserving the order of characters. No, characters may map to the same character and a character may map to itself.
For example,
Given "egg" , "add" return True.
Given "foo" , "bar" return FALSE.
Given "paper" , "title" return True.

Main Topic

Given two strings s and T, determine if they are isomorphic. If the string s can get the string T in the way of character substitution, then s and T are isomorphic. Each occurrence of a character must be replaced by its corresponding character, as well as ensuring that the original order has not changed. Two characters cannot be mapped to the same character, but a character can be mapped to itself.

Thinking of solving problems

"As long as s and T meet one-to-one mapping."
Use a hash table map to maintain the mapping of characters in two strings, and save the mapped values with a set. (S[i], t[i]), if the s[i] key does not appear in the map and T[i] does not appear in the set, it is added to the mapping relationship, the t[i] value has been present, the description is a many-to-one mapping, does not conform to return false. S[i] If it has already occurred, set to S[k], the corresponding map value is t[k]), that is s[i]==s[k], then find s[k] of the corresponding value t[k], if T[I]!=T[K], the existence of a single character two different mappings, two strings are not isomorphic, Returns false, continuing with the next character until the end.

Code Implementation

Algorithm implementation class

ImportJava.Util.HashMap;ImportJava.Util.HashSet;ImportJava.Util.Map;ImportJava.Util.Set; PublicClass Solution { PublicBoolean isisomorphic (StringSStringT) {//Two strings are empty        if(s== NULL &&T== NULL) {return true; }//Only one is empty        Else if(s== NULL ||T== NULL) {return false; }//Two string lengths are 0        Else if(s.Length ()== 0 &&T.Length ()== 0) {return true; }///Two strings are not equal in length        Else if(s.Length ()!=T.Length ()) {return false; }//Save Mapping Relationship        Map<Character, Character> Map = NewHashMap<>(s.Length ());Set<Character> Set = NewHashSet<>(t.Length ());        Char Schar;        Char TChar; for (int i= 0; I<S.Length (); I++) {Schar=S.CharAt (i); TChar=T.CharAt (i);//key is not present, the mapping relationship is saved            if(!Map.ContainsKey (Schar)) {if(Set.Contains (TChar)) {return false; }Else{Map.Put (s.CharAt (i), t.CharAt (i));Set.Add (TChar); }            }//If the key has already appeared            Else{The value of the original key mapping is Map.get (Schar), and now the value to be mapped is TCHAR                //If two values are not equal, the description has been mapped two times, does not conform, returns false                if(Map.Get (Schar)!=TChar) {return false; }            }        }return true; }}

Evaluation Results

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"Leetcode-Interview algorithm classic-java Implementation" "205-isomorphic Strings (isomorphic string)"