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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Grado
Data Structures and Algorithms in Java 6th edition
Institución
Data Structures And Algorithms In Java 6th Edition
Book
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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