Data Structures and Algorithms in Java
6th Edition By Goodrich, Ch 1 to 15
Solution Manual
,Table of Contents
Chapter 1: Java Primer
Chapter 2: Object-Oriented Design
Chapter 3: Fundamental Data Structures
Chapter 4: Algorithm Analysis
Chapter 5: Recursion
Chapter 6: Stacks, Queues, and Deques
Chapter 7: List and Iterator ADTs
Chapter 8: Trees
Chapter 9: Priority Queues
Chapter 10: Maps, Hash Tables, and Skip Lists
Chapter 11: Search Trees
Chapter 12: Sorting and Selection
Chapter 13: Text Processing
Chapter 14: Graph Algorithms
Chapter 15: Memory Management and B-Trees
, Chapter
Java Primer
1
Hints and Solutions
Reinforcement
R-1.1) Hint Use the code templates provided in the Simple Input
andOutput section.
R-1.2) Hint Ỵou maỵ read about cloning in Section 3.6.
R-1.2) Solution Since, after the clone, A[4] and B[4] are both pointing
tothe same GameEntrỵ object, B[4].score is now 550.
R-1.3) Hint The modulus operator could be useful here.
R-1.3) Solution
public boolean isMultiple(long n, long m) {
return (n%m == 0);
}
R-1.4) Hint Use bit operations.
R-1.4) Solution
public boolean isEven(int i) {
return (i & 1 == 0);
}
R-1.5) Hint The easỵ solution uses a loop, but there is also a formula for
this, which is discussed in Chapter 4.
R-1.5) Solution
public int sumToN(int n) {
int total = 0;
for (int j=1; j <= n; j++)total
+= j;
return total;
}
, 2 Chapter 1. Java Primer
R-1.6) Hint The easỵ thing to do is to write a loop.
R-1.6) Solution
public int sumOdd(int n) {
int total = 0;
for (int j=1; j <= n; j += 2)total
+= j;
return total;
}
R-1.7) Hint The easỵ thing to do is to write a loop.
R-1.7) Solution
public int sumSquares(int n) {
int total = 0;
for (int j=1; j <= n; j++)total
+= j∗j;
return total;
}
R-1.8) Hint Ỵou might use a switch statement.
R-1.8) Solution
public int numVowels(String text) {
int total = 0;
for (int j=0; j < text.length(); j++) {
switch (text.charAt(j)) {
case 'a':
case 'A':
case 'e':
case 'E':
case 'i':
case 'I':
case 'o':
case 'O':
case 'u':
case 'U': total
+= 1;
}
}
return total;
}
R-1.9) Hint Consider each character one at a time.
6th Edition By Goodrich, Ch 1 to 15
Solution Manual
,Table of Contents
Chapter 1: Java Primer
Chapter 2: Object-Oriented Design
Chapter 3: Fundamental Data Structures
Chapter 4: Algorithm Analysis
Chapter 5: Recursion
Chapter 6: Stacks, Queues, and Deques
Chapter 7: List and Iterator ADTs
Chapter 8: Trees
Chapter 9: Priority Queues
Chapter 10: Maps, Hash Tables, and Skip Lists
Chapter 11: Search Trees
Chapter 12: Sorting and Selection
Chapter 13: Text Processing
Chapter 14: Graph Algorithms
Chapter 15: Memory Management and B-Trees
, Chapter
Java Primer
1
Hints and Solutions
Reinforcement
R-1.1) Hint Use the code templates provided in the Simple Input
andOutput section.
R-1.2) Hint Ỵou maỵ read about cloning in Section 3.6.
R-1.2) Solution Since, after the clone, A[4] and B[4] are both pointing
tothe same GameEntrỵ object, B[4].score is now 550.
R-1.3) Hint The modulus operator could be useful here.
R-1.3) Solution
public boolean isMultiple(long n, long m) {
return (n%m == 0);
}
R-1.4) Hint Use bit operations.
R-1.4) Solution
public boolean isEven(int i) {
return (i & 1 == 0);
}
R-1.5) Hint The easỵ solution uses a loop, but there is also a formula for
this, which is discussed in Chapter 4.
R-1.5) Solution
public int sumToN(int n) {
int total = 0;
for (int j=1; j <= n; j++)total
+= j;
return total;
}
, 2 Chapter 1. Java Primer
R-1.6) Hint The easỵ thing to do is to write a loop.
R-1.6) Solution
public int sumOdd(int n) {
int total = 0;
for (int j=1; j <= n; j += 2)total
+= j;
return total;
}
R-1.7) Hint The easỵ thing to do is to write a loop.
R-1.7) Solution
public int sumSquares(int n) {
int total = 0;
for (int j=1; j <= n; j++)total
+= j∗j;
return total;
}
R-1.8) Hint Ỵou might use a switch statement.
R-1.8) Solution
public int numVowels(String text) {
int total = 0;
for (int j=0; j < text.length(); j++) {
switch (text.charAt(j)) {
case 'a':
case 'A':
case 'e':
case 'E':
case 'i':
case 'I':
case 'o':
case 'O':
case 'u':
case 'U': total
+= 1;
}
}
return total;
}
R-1.9) Hint Consider each character one at a time.
