（精品课件）第四次chapter4作业1（10.24评讲）.docVIP

P130 E2 E2. Consider a linked stack that includes a method size. This method size requires a loop that moves through the entire stack to count the entries, since the number of entries in the stack is not kept as a separate member in the stack record. Write a method size for a linked stack by using a loop that moves a pointer variable from node to node through the stack. Answer int Stack :: size() const /* Post: Return the number of entries in the Stack. */ { Node *temp = top_node; int count = 0; while (temp != NULL) { temp = temp -next; count++; } return count; } Consider modifying the declaration of a linked stack to make a stack into a structure with two members, the top of the stack and a counter giving its size. What changes will need to be made to the other methods for linked stacks? Discuss the advantages and disadvantages of this modi?cation compared to the original implementation of linked stacks. Answer 构造函数需添加将count初始化为0的语句。 empty函数和size函数只要根据count返回相应结果。 成功的push操作应使count增1 成功的pop操作应使count减1 优点：可使链栈类与顺序栈类一致。 减少size方法的运行时间。 缺点：增加额外的变量和额外的代码。 是否添加count，取决于size方法的调用频率。 P137 4.3 E2 E2. What is wrong with the following attempt to use the copy constructor to implement the overloaded assignment operator for a linked Stack? void Stack :: operator = (const Stack original) { Stack new_copy(original); top_node = new_copy.top_node; } How can we modify this code to give a correct implementation? Answer The assignment top_node = new_copy.top_node; leaves the former nodes of the left hand side of the assignment operator as garbage.应该回收的空间没回收，结果丢了变成垃圾！ Moreover, when the function ends, the statically allocated Stack new_copy will be destructed, this will destroy the newly assigned left hand side of the assignment operator. 把需要赋值不该回收的空间回收了。 We can avoid both problems by swapping the pointers top_node and new_copy.top_node as in the following. //用书上的方法也可以 void Stack :: operator = (const Stack original) /* Post: The Stack is reset as a copy of Stack original. */ { Stack n