Data Structures And Algorithms In Java 6th Edition
Data Structures and Algorithms in Java 6th edition
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Solution Manual for Data Structures and Algorithms in Java 6th edition by Michael T. Goodrich || A+
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Data Structures and Algorithms in Java 6th edition
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Data Structures And Algorithms In Java 6th Edition
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Data Structures and Algorithms in Java, 6th Edition
Solution Manual for Data Structures and Algorithms in Java 6th edition by Michael T. Goodrich || A+
Chapter
1
Java Primer
Hints and Solutions
Reinforcement
R-1.1) Hint Use the code templates provided in the Simple Input and
Output section.
R-1.2) Hint You may read about cloning in Section 3...
Solution Manual for Data Structures and Algorithms in Java 6th edition by Michael T. Goodrich
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Solutions Manual for
Data Structures and
Algorithms in Java, 6e
Michael Goodrich,
Roberto Tamassia (All
Chapters)
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, Chapter
1 Java Primer
Hints and Solutions
Reinforcement
R-1.1) Hint Use the code templates provided in the Simple Input and
Output section.
R-1.2) Hint You may read about cloning in Section 3.6.
R-1.2) Solution Since, after the clone, A[4] and B[4] are both pointing to
the same GameEntry object, B[4].score is now 550.
R-1.3) Hint The modulus operator could be useful here.
R-1.3) Solution
public boolean isMultiple(long n, long m) {
return (n%m == 0);
}
R-1.4) Hint Use bit operations.
R-1.4) Solution
public boolean isEven(int i) {
return (i & 1 == 0);
}
R-1.5) Hint The easy solution uses a loop, but there is also a formula for
this, which is discussed in Chapter 4.
R-1.5) Solution
public int sumToN(int n) {
int total = 0;
for (int j=1; j <= n; j++)
total += j;
return total;
}
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,2 Chapter 1. Java Primer
R-1.6) Hint The easy thing to do is to write a loop.
R-1.6) Solution
public int sumOdd(int n) {
int total = 0;
for (int j=1; j <= n; j += 2)
total += j;
return total;
}
R-1.7) Hint The easy thing to do is to write a loop.
R-1.7) Solution
public int sumSquares(int n) {
int total = 0;
for (int j=1; j <= n; j++)
total += j∗j;
return total;
}
R-1.8) Hint You might use a switch statement.
R-1.8) Solution
public int numVowels(String text) {
int total = 0;
for (int j=0; j < text.length( ); j++) {
switch (text.charAt(j)) {
case 'a':
case 'A':
case 'e':
case 'E':
case 'i':
case 'I':
case 'o':
case 'O':
case 'u':
case 'U':
total += 1;
}
}
return total;
}
R-1.9) Hint Consider each character one at a time.
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, 3
R-1.10) Hint Consider using get and set methods for accessing and mod-
ifying the values.
R-1.11) Hint The traditional way to do this is to use setFoo methods,
where Foo is the value to be modified.
R-1.11) Solution
public void setLimit(int lim) {
limit = lim;
}
R-1.12) Hint Use a conditional statement.
R-1.12) Solution
public void makePayment(double amount) {
if (amount > 0)
balance −= amount;
}
R-1.13) Hint Try to make wallet[1] go over its limit.
R-1.13) Solution
for (int val=1; val <= 58; val++) {
wallet[0].charge(3∗val);
wallet[1].charge(2∗val);
wallet[2].charge(val);
}
This change will cause wallet[1] to attempt to go over its limit.
Creativity
C-1.14) Hint The Java method does not need to be passed the value of n
as an argument.
C-1.15) Hint Note that the Java program has a lot more syntax require-
ments.
C-1.16) Hint Create an enum type of all operators, including =, and use
an array of these types in a switch statement nested inside for-loops to try
all possibilities.
C-1.17) Hint Note that at least one of the numbers in the pair must be
even.
C-1.17) Solution
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