Lintcode Easy title: Unique characters to determine if a string has no duplicate characters

Topic:

Determine if a string has no repeating characters

Implements an algorithm that determines whether characters in a string are unique occurrences

Sample Example

Give "ABC", Return True

Give "AaB", return False

Challenge

If you don't use extra storage space, how does your algorithm change?

Solving:

Defining a collection is the simplest.

Java Program:

 Public classSolution {/**     * @paramstr:a String *@return: A Boolean*/     Public BooleanIsUnique (String str) {//Write your code hereTreeSet set =NewTreeSet ();  for(intI=0;i<str.length (); i++)            if(Set.add (Str.charat (i)) = =false)                return false; return true; }}

View Code

Total time: 2209 ms

Java Program:

 Public classSolution {/**     * @paramstr:a String *@return: A Boolean*/     Public BooleanIsUnique (String str) {//Write your code here         for(intI=0;i<str.length (); i++){             for(intJ=i+1;j<str.length (); j + +){                if(Str.charat (i) = =Str.charat (j))return false; }        }        return true; }}

View Code

Total time: 1095 ms

This should not count as extra storage space, Time complexity O (n2)

Python program:

Using dictionaries

classSolution:#@param s:a String    #@return: A Boolean    defIsUnique (self, str):#Write your code hereD = {}         forSinchStr:ifS not inchD:d[s]= 1Else:                returnFalsereturnTrue

View Code

Total time: 255 ms

Lintcode Easy title: Unique characters to determine if a string has no duplicate characters