1. Problem Name : helloName
Solution:
Hint: The code
Return
"xyz" + name;
public String
helloName(String name)
{
return("Hello "+name+"!");
}
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2. Problem Name : makeAbba
Solution:
Hint:
e.g. "Hi" and "Bye" returns "HiByeByeHi". Return
(a+b+b+a)
public String
makeAbba(String a, String b)
{
return (a+b+b+a);
}
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3. Problem Name : makeTags
Solution:
Hint:
Given tag and word strings, create the HTML string with tags around the word,
e.g. "<i>Yay</i>". ‘i’ should be replaced by
whatever is given in tag string and Yay should be replaced by whatever is given
in word string.
public String
makeTags(String tag, String word)
{
String starttag="<"+tag+">";
String endtag="</"+tag+">";
return(starttag+word+endtag);
}
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4. Problem Name : makeOutWord
Solution:
Hint:
Need to enclose the words in the given
parenthesis and return the string. The
out word contains e.g. “<<>>” the first two characters represents
our start tag and last two characters of out represents end tag. Return Start
public String
makeOutWord(String out, String word)
{
int len=out.length();// length of out
string
String
starttag=out.substring(0,2); // to take << from out string
String endtag=out.substring(len-2,len); // to take
>> from out string
return(starttag+word+endtag);
}
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5. Problem Name : extraEnd
Solution:
Hint: return a new string made of 3 copies of
the last 2 chars of the original string.
public String
extraEnd(String str)
{
String end =
str.substring(str.length()-2);
return end + end + end;
}
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6. Problem Name : firstTwo
Solution:
Hint: return the string made of its first two chars, so the String
"Hello" yields "He".
Constraint:
If the string is shorter than length 2, return whatever there is, so
"X" yields "X", and the empty string ""
public String
firstTwo(String str)
{
if (str.length() >= 2)
{
return str.substring(0, 2);
}
else
{
return str;
}
}
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7. Problem Name : firstHalf
Solution:
Hint: Given a string
of even length, return the first half. So the string "WooHoo" yields
"Woo"
public String
firstHalf(String str)
{
int len=str.length();
if(len%2==0)
{
int half=len/2;
return(str.substring(0,half));
}
else
{
return " ";
}
}
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8. Problem Name : withoutEnd
Solution:
Hint: Given a string, return
a version without the first and last char, so "Hello" yields
"ell".
Constraint:
The string length will be at least 2.
public String
withoutEnd(String str)
{
return (str.substring(1,str.length()-1));
}
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9. Problem Name : comboString
Solution:
Hint: Given 2 strings, a and
b, return a string of the form short+long+short, with the shorter string on the
outside and the longer string on the inside.
Constraint:
The strings will not be the same length, but they may be empty (length 0)
public String
comboString(String a, String b)
{
if(a.length()>b.length())
return(b+a+b);
else
return(a+b+a);
}
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10. Problem Name : nonStart
Solution:
Hint: Given 2 strings,
return their concatenation, except omit the first char of each
Constraint:
The strings will be at least length 1.
public String
nonStart(String a, String b)
{
return(a.substring(1,a.length())+b.substring(1,b.length()));
}
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11. Problem Name : left2
Solution:
Hint:
Given a string, return a "rotated left 2" version where the first 2
chars are moved to the end
Constraint:
The string length will be at least 2.
public String
left2(String str)
{
return str.substring(2) + str.substring(0,
2);
//substring(2) extracts the string
from index 2 through the end. So we put that first, followed by substring(0,
2)
}
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12. Problem Name : right2
Solution:
Hint: Given a string, return a
"rotated right 2" version where the last 2 chars are moved to the
start.
Constraint:
The string length will be at least 2.
public String
right2(String str)
{
int len=str.length();
if(len>2)
{
return(str.substring(len-2,len)+str.substring(0,len-2));
}
else
{
return str;
}
}
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13. Problem Name : theEnd
Solution:
Hint: Given a string, return a
string length 1 from its front, unless front is false, in which case return a
string length 1 from its back.
Constraint:
The string will be non-empty.
public String
theEnd(String str, boolean front)
{
if(front)
return(str.substring(0,1));
else
return
(str.substring(str.length()-1,str.length()));
}
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14. Problem Name : withouEnd2
Solution:
Hint: Given a string, return a version without both the first and last
char of the string.
Constraint:
The string may be any length, including 0.
public String
withouEnd2(String str)
{
if(str.length()>2)
{
return(str.substring(1,str.length()-1));
}
else
return "";
}
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15. Problem Name : middleTwo
Solution:
Hint: Given a string of even length, return a string made of the middle
two chars, so the string "string" yields "ri".
Constraint:
The string length will be at least 2.
public String
middleTwo(String str)
{
if(str.length()>2)
{
int middle=str.length()/2;
return(str.substring(middle-1,middle+1));
}
else
{
return str;
}
return result;
}
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16. Problem Name : endsLy
Solution:
Hint:
Given a string, return true if it ends in "ly".
public boolean endsLy(String str)
{
int len=str.length();
if(len>=2)
{
if(str.substring(len-2,len).equals("ly"))
return true;
else
return false;
}
else
{
return false;
}
}
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17. Problem Name : nTwice
Solution:
Hint: return a string made of the first and last n chars from the string.
public String nTwice(String
str, int n)
{
return
(str.substring(0,n)+str.substring(str.length()-n,str.length()));
}
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18. Problem Name : twoChar
Solution:
Hint: return a string length 2 starting at the given index. If the index is too big or too small to define a string length 2, use the first 2 chars.
Constraint: The string length will be at least 2.
public String
twoChar(String str, int index)
{
int len=str.length();
if(index>=0 && index+2<=len)
return(str.substring(index,index+2));
else
return (str.substring(0,2));
}
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Please check the my posts for solutions to the remaining problems of Java String 1 section.
