Solution Manual for Data Structures and Algorithms in Java 6th edition by Michael T. Goodrich || A+

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

, Chapter


1 Java Primer XM




Hints and Solutions XM XM




Reinforcement
R-
1.1) Hint Use the code templates provided in the Simple Input an
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d 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 th
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e 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 XM




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




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




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




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




public int sumToN(int n) {
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int total = 0;
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for (int j=1; j <= n; j++) tota
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l += j; XM XM




return total; X M




}

,2 Chapter 1. Java Prime XM X M XM




r
R-1.6) Hint The easy thing to do is to write a loop.
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R-1.6) Solution
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public int sumOdd(int n) {
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int total = 0;
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for (int j=1; j <= n; j += 2) t
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otal += j; XM XM




return total; X M




}
R-1.7) Hint The easy thing to do is to write a loop.
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R-1.7) Solution
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public int sumSquares(int n) {
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int total = 0;
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for (int j=1; j <= n; j++) tota
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l += j∗j;
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return total; X M




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




case 'a': XM




case 'A': XM




case 'e': XM




case 'E': XM




case 'i': XM




case 'I': XM




case 'o': XM




case 'O': XM




case 'u': XM




case 'U': tot XM XM




al += 1; XM XM




}
}
return total; X M




}
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 XM




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




limit = lim; XM XM




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




public void makePayment(double amount) {
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if (amount > 0) balanc
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e − = amount; XM XM




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




for (int val=1; val <= 58; val++) { wallet[0].
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charge(3∗val); wallet[1].charge(2∗val); w XM XM




allet[2].charge(val);
}
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 a
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n array of these types in a switch statement nested inside for-
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loops to try all possibilities.
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C-
1.17) Hint Note that at least one of the numbers in the pair must be ev
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en.
C-1.17) Solution XM