Polynomial add(const Polynomial& p1, const Polynomial& p2) { Node* curr1 = p1.head; Node* curr2 = p2.head; Polynomial result; while (curr1 && curr2) { if (curr1->exp > curr2->exp) { result.insert(curr1->coeff, curr1->exp); curr1 = curr1->next; } else if (curr1->exp < curr2->exp) { result.insert(curr2->coeff, curr2->exp); curr2 = curr2->next; } else { int sum = curr1->coeff + curr2->coeff; result.insert(sum, curr1->exp); curr1 = curr1->next; curr2 = curr2->next; } } while (curr1) { result.insert(curr1->coeff, curr1->exp); curr1 = curr1->next; } while (curr2) { result.insert(curr2->coeff, curr2->exp); curr2 = curr2->next; } return result; }函数功能
时间: 2024-03-15 08:43:42 浏览: 70
该函数实现了两个多项式相加的功能。它接受两个多项式对象作为输入,将它们的系数和指数相加,并将结果存储在一个新的多项式对象中,最后返回该新对象。该函数使用了链表来存储多项式的系数和指数,通过遍历两个多项式的链表进行相加操作。具体实现过程是,先初始化一个新的多项式对象,然后遍历两个输入多项式的链表,如果两个链表节点的指数相同,则将它们的系数相加,并插入到结果多项式的链表中;如果一个链表的节点的指数大于另一个链表的节点的指数,则将较大指数的节点插入到结果多项式的链表中,并将指向该链表的指针向后移动一位;如果一个链表的节点的指数小于另一个链表的节点的指数,则将较小指数的节点插入到结果多项式的链表中,并将指向该链表的指针向后移动一位。最后,如果有一个多项式的链表还有剩余节点,则将它们插入到结果多项式的链表中。
相关问题
Polynomial subtract(const Polynomial& p1, const Polynomial& p2) { Node* curr1 = p1.head; Node* curr2 = p2.head; Polynomial result; while (curr1 && curr2) { if (curr1->exp > curr2->exp) { result.insert(curr1->coeff, curr1->exp); curr1 = curr1->next; } else
if (curr1->exp < curr2->exp) { result.insert(-curr2->coeff, curr2->exp); curr2 = curr2->next; } else { double tempCoeff = curr1->coeff - curr2->coeff; if (tempCoeff != 0) { result.insert(tempCoeff, curr1->exp); } curr1 = curr1->next; curr2 = curr2->next; } } while (curr1) { result.insert(curr1->coeff, curr1->exp); curr1 = curr1->next; } while (curr2) { result.insert(-curr2->coeff, curr2->exp); curr2 = curr2->next; } return result; }
以上是一个多项式相减的函数实现,你有什么问题吗?
Polynomial add(const Polynomial& p1, const Polynomial& p2) { Node* curr1 = p1.head; Node* curr2 = p2.head; Polynomial result; while (curr1 && curr2) { if (curr1->exp > curr2->exp) { result.insert(curr1->coeff, curr1->exp); curr1 = curr1->next; } else if (curr1->exp < curr2->exp) { result.insert(curr2->coeff, curr2->exp); curr2 = curr2->next; } else { int sum = curr1->coeff + curr2->coeff; result.insert(sum, curr1->exp); curr1 = curr1->next; curr2 = curr2->next; } } while (curr1) { result.insert(curr1->coeff, curr1->exp); curr1 = curr1->next; } while (curr2) { result.insert(curr2->coeff, curr2->exp); curr2 = curr2->next; } return result; }
这段代码实现了两个多项式相加的函数 add,它使用了之前实现的多项式类。具体实现是通过遍历两个多项式的链表,将它们的系数相加并插入到结果多项式中。如果某个多项式已经遍历完,则将剩余的节点插入到结果多项式中。最后返回结果多项式。
需要注意的是,这个函数返回的是一个新的多项式,而不是修改原有的多项式。因此,可以在调用该函数时传入 const 引用,以避免不必要的复制。