利用链表实现两个稀疏多项式相加。

代码

#include <iostream>
using namespace std;

// 定义多项式项结构体
typedef struct {
    int x;  // 系数
    int y;  // 指数
} Elemtype;

// 定义链表节点结构体
typedef struct Node {
    Elemtype data;
    struct Node* next;
} *LinkList, Node;

// 创建多项式链表，按照乱序输入顺序输出
void Great_LinkList(LinkList& L) {
    int n;
    Node* s, * pre, * p;
    cout << "请输入多项式的项数：";
    cin >> n;
    L = new Node;  // 建立一个带头结点的单链表
    L->next = NULL;
    for (int i = 1; i <= n; i++) {
        s = new Node;
        cout << "请输入第" << i << "项的系数和指数：";
        cin >> s->data.x >> s->data.y;//注意此步在for循环内部，每次都需要从头开始找
        pre = L; p = L->next;
        while (p && p->data.y < s->data.y) {
            pre = p;
            p = p->next;
        }
        s->next = p;
        pre->next = s;
    }
}

// 显示多项式
void Show(LinkList L) {
    Node* p;
    p = L->next;
    cout << "f(x)=";
    while (p) {
        if (p->data.x != 0) {
            if (p->data.x > 0) {
                if (p->data.y)
                    cout << "+" << p->data.x << "x^" << p->data.y;
                else
                    cout << p->data.x;
            }
            else {
                if (p->data.y)
                    cout << p->data.x << "x^" << p->data.y;
                else
                    cout << p->data.x;
            }
        }
        p = p->next;
    }
}

// 多项式相加
void Add_List(LinkList& LA, LinkList& LB) {
    Node* pa, * pb, * pc, * r;
    pc = LA;  // 复制 LA 链表作为 LC 的初始值
    pa = LA->next; pb = LB->next;
    while (pa && pb) {
        if (pa->data.y == pb->data.y) {
            pa->data.x += pb->data.x;  // 相加系数
            if (pa->data.x != 0) {  // 系数相加不为 0
                pc->next = pa;
                pc = pa;
                pa = pa->next;
                r = pb; pb = pb->next; delete r;  // 跳过当前项

            }
            else {  // 系数为 0，删除当前项
                r = pa; pa = pa->next; delete r;
                r = pb; pb = pb->next; delete r;
            }
        }
        else if (pa->data.y < pb->data.y) {
            pc->next = pa;
            pc = pa;
            pa = pa->next;
        }
        else {
            pc->next = pb;
            pc = pb;
            pb = pb->next;
        }
    }
    pc->next = pa ? pa : pb;  // 连接剩余项
    Show(LA);

}

int main() {
    int n;
    LinkList L1, L2, LC;
    cout << "开始创建第一个多项式：" << endl;
    Great_LinkList(L1);  // 创建第一个多项式
    cout << "开始创建第二个多项式：" << endl;
    Great_LinkList(L2);  // 创建第二个多项式

    cout << "你创建的第一个多项式为：";
    Show(L1);
    cout << endl;

    cout << "你创建的第二个多项式为：";
    Show(L2);
    cout << endl;

    cout << "两个多项式相加的结果为：";
    Add_List(L1, L2);
    cout << endl;
    return 0;
}

void Add_List(LinkList & LA, LinkList & LB, LinkList & LC) {
		Node* pa, * pb, * pc;
		LC = LA, detete LB;
		pa = LA->next; pb = LB->next;
		pc = LC->next;
		while (pa && pb) {
			if (pa->data.y == pb->data.y) {
				sum = pa->data.y + pb->data.y;
				if (sum) {  //系数和不为0
					pa->data.y = sum;
					pc->next = pa;
					pc = pa;
					pa = pa->next;
					r = pb; pb = pb->next; delete pb;
				}
				else {   //系数相加为0
					r = pa; pa = pa->next; delete pa;
					r = pb; pb = pb->next; delete pb;
				}
			}
			else if (pa->data.y < pb->data.y) {
				pc->next = pa;
				pc = pa;
				pa = pa->next;
			}
			else {
				pc->next = pb;
				pc = pb;
				pb = pb->next;
			
			}

			pc->next = pa ? pa : pb;
			Show(LC);

 运行结果

下面是我最初写的多项式相加的代码，传了一个LC过来，最后运行出现了错误。  

void Add_List(LinkList & LA, LinkList & LB, LinkList & LC) {
		Node* pa, * pb, * pc;
		LC = LA, detete LB;
		pa = LA->next; pb = LB->next;
		pc = LC->next;
		while (pa && pb) {
			if (pa->data.y == pb->data.y) {
				sum = pa->data.y + pb->data.y;
				if (sum) {  //系数和不为0
					pa->data.y = sum;
					pc->next = pa;
					pc = pa;
					pa = pa->next;
					r = pb; pb = pb->next; delete pb;
				}
				else {   //系数相加为0
					r = pa; pa = pa->next; delete pa;
					r = pb; pb = pb->next; delete pb;
				}
			}
			else if (pa->data.y < pb->data.y) {
				pc->next = pa;
				pc = pa;
				pa = pa->next;
			}
			else {
				pc->next = pb;
				pc = pb;
				pb = pb->next;
			
			}

			pc->next = pa ? pa : pb;
			Show(LC);