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

Hints and Solutions Reinforcement R-1.1) Hint Use the code templates provided in the Simple Input and Output section. R-1.2) Hint You may read about cloning in Section 3.6. R-1.2) Solution Since, after the clone, A[4] and B[4] are both pointing to the same GameEntry 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 easy 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; } Chapter 1. Java Primer 2 R-1.6) Hint The easy 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 easy 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 You might use a switch statement. R-1.8) Solution public int numVowels(String text) { int total = 0; for (int j=0; j < h(); j++) { switch (At(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.