242. Valid Anagram

Topic:

Given, Strings s and T, write a function to determine if it is a anagram of s.

For example,
s = "Anagram", t = "Nagaram", return true.
s = "Rat", t = "Car", return false.

Note:
Assume the string contains only lowercase alphabets.

Follow up:
What if the inputs contain Unicode characters? How would adapt your solution to such case?

Links: http://leetcode.com/problems/valid-anagram/

Exercises

Verify the anagram. There are many solutions to this problem.

The simplest is to sort the array after the comparison.

Time Complexity-o (NLOGN), Space complexity-o (n)

 Public classSolution { Public BooleanIsanagram (string s, String t) {if(s = =NULL|| t = =NULL|| S.length ()! =t.length ())return false; if(s.length () = = 0)            return true; Char[] SArr =S.tochararray (); Char[] TArr =T.tochararray ();        Arrays.sort (SARR);                Arrays.sort (TARR); returnarrays.equals (SARR, TARR); }}

We can also use HashMap, traverse s to save the elements in the map, and then traverse t to check if there are exactly the same elements and quantities as s.

Time Complexity-o (n), Space complexity-o (1). But the actual speed of operation is rather slow.

 Public classSolution { Public BooleanIsanagram (string s, String t) {if(s = =NULL|| t = =NULL|| S.length ()! =t.length ())return false; if(s.length () = = 0)            return true; Map<character, integer> map =NewHashmap<>();  for(inti = 0; I < s.length (); i++) {            Charc =S.charat (i); if(Map.containskey (c)) Map.put (c, Map.get (c)-W); ElseMap.put (c,1); }                 for(inti = 0; I < t.length (); i++) {            Charc =T.charat (i); if(!Map.containskey (c))return false; Else {                if(Map.get (c) = = 0)                    return false; Map.put (c, Map.get (c)-1); }        }                return true; }}

can also use bitmap, because the title description only contains lowercase letters, so we can build a length of 26 bitmap, while traversing s and T, increment bitmap[char S-' a '], decrement bitmap[char t-' a '], Finally check if each bit is 0, yes returns true, otherwise false is returned.

 Public classSolution { Public BooleanIsanagram (string s, String t) {if(s = =NULL|| t = =NULL|| S.length ()! =t.length ())return false; if(s.length () = = 0)            return true; int[] BitMap =New int[26];  for(inti = 0; I < s.length (); i++) {Bitmap[s.charat (i)-' A ']++; Bitmap[t.charat (i)-' A ']--; }                 for(intI:bitmap)if(I! = 0)                return false; return true; }}

Reference:

Https://leetcode.com/discuss/57903/java-solution-using-sort

Https://leetcode.com/discuss/49399/accepted-java-o-n-solution-in-5-lines

Https://leetcode.com/discuss/61456/my-java-solution-8ms

Https://leetcode.com/discuss/49795/share-my-java-solution

Https://leetcode.com/discuss/50631/jave-simple-and-efficient-solution

242. Valid Anagram