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)

, Chapter


1 Java Primer xz




Hints and Solutions xz xz




Reinforcement
R-
1.1) Hint Use the code templates provided in the Simple Input and O
x z xz x z x z x z x z x z x z x z x z x z xz




utput section. xz




R-1.2) Hint You may read about cloning in Section 3.6.
xz xz xz xz xz xz xz xz xz




R-
1.2) Solution Since, after the clone, A[4] and B[4] are both pointing to the sa
xz xz xz xz xz xz xz xz xz xz xz xz xz xz




me GameEntryobject, B[4].scoreis now 550.
xz xz xz xz xz xz




R-1.3) Hint The modulus operator could be useful here.
xz xz xz xz xz xz xz xz




R-1.3) Solution xz




public boolean isMultiple(long n, long m) {
xz xz xz xz xz xz




return (n%m == 0); xz xz xz




}
R-1.4) Hint Use bit operations.
xz xz xz xz




R-1.4) Solution xz




public boolean isEven(int i) {
xz xz xz xz




return (i & 1 == 0); xz xz xz xz xz




}
R-
1.5) Hint The easy solution uses a loop, but there is also a formula for this,
xz xz xz xz xz xz xz xz xz xz xz xz xz xz xz xz




which is discussed in Chapter 4.
xz xz xz xz xz




R-1.5) Solution xz




public int sumToN(int n) {
xz xz xz xz




int total = 0;
xz xz xz




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




+= j; xz




return total; xz




}

,2 Chapter 1. Java Primer
xz x z xz




R-1.6) Hint The easy thing to do is to write a loop.
xz xz xz xz xz xz xz xz xz xz xz




R-1.6) Solution xz




public int sumOdd(int n) {
xz xz xz xz




int total = 0;
xz xz xz




for (int j=1; j <= n; j += 2) tot
xz xz xz xz xz xz xz xz xz




al += j;
xz xz




return total; xz




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




R-1.7) Solution xz




public int sumSquares(int n) {
xz xz xz xz




int total = 0;
xz xz xz




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




+= j∗j; xz




return total; xz




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




R-1.8) Solution xz




public int numVowels(String text) {
xz xz xz xz




int total = 0;
xz xz xz




for (int j=0; j < text.length(); j++) {
xz xz xz xz xz xz xz




switch (text.charAt(j)) { xz xz




case 'a': xz




case 'A': xz




case 'e': xz




case 'E': xz




case 'i': xz




case 'I': xz




case 'o': xz




case 'O': xz




case 'u': xz




case 'U': tota xz xz




l+=1; zx xz




}
}
return total; xz




}
R-1.9) Hint Consider each character one at a time.
xz xz xz xz xz xz xz xz

, 3
R-1.10) Hint Consider using get and set methods for accessing and mod-
xz xz xz xz xz xz xz xz xz xz xz




ifying the values.
xz xz xz




R-
1.11) Hint The traditional way to do this is to use setFoo methods, wh
x z xz xz x z x z x z x z x z x z xz x z x z xz




ere Foo is the value to be modified.
xz xz xz xz xz xz xz




R-1.11) Solution xz




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




limit = lim; xz xz




}
R-1.12) Hint Use a conditional statement. xz xz xz xz xz




R-1.12) Solution xz




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




if (amount > 0) balance
xz xz xz xz




−= amount; xz xz




}

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




R-1.13) Solution xz




for (int val=1; val <= 58; val++) { wallet[0].c
xz xz xz xz xz xz xz xz




harge(3∗val); wallet[1].charge(2∗val); wall xz xz




et[2].charge(val);
}
This change will cause wallet[1] to attempt to go over its limit.
xz xz xz xz xz xz xz xz xz xz xz




Creativity
C-1.14) Hint The Java method does not need to be passed the value of n
xz xz xz xz xz xz xz xz xz xz xz xz xz xz




as an argument.
xz xz




C-1.15) Hint Note that the Java program has a lot more syntax require-
xz xz xz xz xz xz xz xz xz xz xz xz




ments.
xz




C-
1.16) Hint Create an enum type of all operators, including =, and use an arr
xz xz xz xz xz xz xz xz xz xz xz xz xz xz




ay of these types in a switch statement nested inside for-
xz xz xz xz xz xz xz xz xz xz




loops to try all possibilities.
xz xz xz xz




C-
1.17) Hint Note that at least one of the numbers in the pair must be even.
xz xz xz xz xz xz xz xz xz xz xz xz xz xz xz




C-1.17) Solution xz