Solutions Manual for Data Structures and Algorithms in Java 6th Edition by Michael Goodrich & Roberto Tamassia | Complete All Chapters

Solutions Manual for k k




Data Structures and
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Algorithms in Java, 6e
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Michael Goodrich,
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k RobertoTamassia(All k k




Chapters)
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, Chapter


1 Java Primer k




Hints and Solutions k k




Reinforcement
R-1.1) Hint Use the code templates provided in the Simple Input and
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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 the
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same GameEntryobject, B[4].scoreis 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 k




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




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




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




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




public int sumToN(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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k+= j; k




return total; k




}

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




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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ktotal += j; k k




return total; k




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




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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k+= j∗j; k




return total; k




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




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)) { k k




case 'a': k




case 'A': k




case 'e': k




case 'E': k




case 'i': k




case 'I': k




case 'o': k




case 'O': k




case 'u': k




case 'U': k




total+=1;
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}
}
return total; k




}
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 k




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




limit = lim; k k




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




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




if (amount > 0) balance
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−= amount; k k




}

R-1.13) Hint Trytomakewallet[1]go overits limit.
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R-1.13) Solution k




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 an
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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 even.
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C-1.17) Solution k