Full Solutions Manual – Data Structures and Algorithms in Java (6th Edition, Goodrich & Tamassia) | Latest 2025 Update | All Chapters

Solutions Manual for
Data Structures and
Algorithms in Java, 6e
Michael Goodrich,
Roberto Tamassia (All
Chapters)

, Chapter


1 Java Primer i




Hints and Solutions i i




Reinforcement
R-1.1) Hint Use the code templates provided in the Simple Input
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and Output section.
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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 to
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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 i




public boolean isMultiple(long n, long m) {
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return (n%m == 0); i i i


}
R-1.4) Hint Use bit operations.
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R-1.4) Solution i




public boolean isEven(int i) {
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return (i & 1 == 0); i i i i i


}
R-1.5) Hint The easy solution uses a loop, but there is also a formula for
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this, which is discussed in Chapter 4.
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R-1.5) Solution i




public int sumToN(int n) {
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int total = 0;i i i


for (int j=1; j <= n; j++) total
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+= j;
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return total; i


}

,2 Chapter 1. Java Primer
R-1.6) Hint The easy thing to do is to write a loop.
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R-1.6) Solution i




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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itotal += j; i i


return total; i


}
R-1.7) Hint The easy thing to do is to write a loop.
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R-1.7) Solution i




public int sumSquares(int n) {
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int total = 0;
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for (int j=1; j <= n; j++) total
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i+= j∗j; i


return total; i


}
R-1.8) Hint You might use a switch statement.
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R-1.8) Solution i




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)) { i i


case 'a': i


case 'A': i


case 'e': i


case 'E': i


case 'i': i


case 'I': i


case 'o': i


case 'O': i


case 'u': i


case 'U': i


total+= 1;
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}
}
return total; i


}
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 methods,
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where Foo is the value to be modified.
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R-1.11) Solution i




public void setLimit(int lim) { i i i i


limit = lim; i i


}
R-1.12) Hint Use a conditional statement.
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R-1.12) Solution i




public void makePayment(double amount) { i i i i


if (amount > 0) i i i


balance −= amount; i i i


}

R-1.13) Hint Try to make wallet[1] go over its limit.
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R-1.13) Solution i




for (int val=1; val <= 58; val++) {
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wallet[0].charge(3∗val);
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wallet[1].charge(2∗val);
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wallet[2].charge(val);
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}
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 require-
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ments.
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C-1.16) Hint Create an enum type of all operators, including =, and use
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an array of these types in a switch statement nested inside for-loops to try all
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possibilities.
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C-1.17) Hint Note that at least one of the numbers in the pair must be
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even.
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C-1.17) Solution i