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Data Str &amp; Algorithm HW Solutions 30

# Data Str &amp; Algorithm HW Solutions 30 - 4.17 int...

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30 Chap. 4 Lists, Stacks, and Queues f.val += val; s.push(f); } else { // switch order to evaluate 2nd operand FIBobj temp; temp.val = val; temp.op = OUT; S.push (f); S.push (temp); } } } return val; // Correct result should be in val now } 4.16 The stack-based version will be similar to the answer for problem 4.15, so I will not repeat it here. The recursive version is as follows. int recur(int n) { if (n == 1) return 1; return recur((n+1)/2) + recur(n/2) + n;
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Unformatted text preview: } 4.17 int GCD1(int n, int m) { if (n < m) swap(n, m); while ((n % m) != 0) { n = n % m; swap(m, n); } return m; } int GCD2(int n, int m) { if (n < m) swap(n, m); if ((n % m) == 0) return m; return GCD2(m, n % m); } 4.18 void reverse(Queue& Q, Stack& S) { ELEM X; while (!Q.isEmpty()) { X = Q.dequeue(); S.push(X); } while (!S.isEmpty()) { X = S.pop(); Q.enqueue(X);...
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