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