数据结构实验2：一元多项式的加法

实现一元多项式的加法（需要编写可运行程序）要求：
1、通过键盘随机输入两个多项式P（x）和Q（x）的内容。
2、输出结果要有P（x）、Q（x）以及他们的和。
3、输入输出多项式的格式可自行定义。

#include 
#include 
#include 
struct NODE{
    int idx;
    int exp;
    struct NODE* next;
};

void QuickSort(struct NODE a[], int left, int right)
{
    //快排，从大到小
    if(left >= right)   return ;
    int i = left, j = right, key = a[left].exp;
    struct NODE tp=a[left];
    while(i < j) {
        while(i < j && key >= a[j].exp) j--;
        a[i] = a[j];
        while(i < j && key <= a[i].exp) i++;
        a[j] = a[i];
    }
    a[i] = tp;
    QuickSort(a, left, i - 1);
    QuickSort(a, i + 1, right);
}
void Attach(int idx,int exp, struct NODE** pRear){
    struct NODE* P;
    P = (struct NODE*)malloc(sizeof(struct NODE*));
    P->exp=exp;
    P->idx=idx;
    P->next=NULL;
    (*pRear)->next=P;
    *pRear = P;
}

struct NODE* build(){
    struct NODE* P;
    struct NODE* Rear;
    struct NODE* t;
    int n,a,b;
    scanf("%d",&n);
    P=(struct NODE*)malloc(sizeof(struct NODE));
    P->next=NULL;
    Rear = P;
    while(n--){
        scanf("%d%d",&a,&b);
        Attach(a,b,&Rear);
    }
    //删除头结点
    t=P;
    P=P->next;
    free(t);
    return P;
}
void mysotr(struct NODE* head){
    //先把链表的内容存放到数组中（时间为O(n)），然后对那个数组进行从大到小排序nlog(n)
    int i=0,len=0;
    struct NODE *p= (struct NODE *) head;
    while(p!=NULL){
        len++;
        p=p->next;
    }
    struct NODE a[len+2];
    p= (struct NODE *) head;
    for(i=0;p!=NULL;p=p->next,i++){
        a[i].idx=p->idx;
        a[i].exp=p->exp;
    }
    QuickSort(a,0,len-1);
    p= (struct NODE *) head;
    for(i=0;p!=NULL;p=p->next,i++){
        p->idx=a[i].idx;
        p->exp=a[i].exp;
    }
}
void myprint(struct NODE* head){

    if (!head) {
        printf("0 0n");
        return;
    }
    struct NODE *p=head;
    while(p!=NULL){
        if(p!=head) printf(" ");
        printf("%d %d",p->idx,p->exp);
        p = p->next;
    }
    printf("n");
}
struct NODE* Add(struct NODE* p1,struct NODE* p2){
    struct NODE* PS;
    struct NODE* rear;
    struct NODE* temp;
    int sum=0;
    PS = (struct NODE*) malloc(sizeof(struct NODE));
    PS->next=NULL;
    rear = PS;
    while(p1!=NULL&&p2!=NULL){
        if(p1->exp==p2->exp){
            sum = p1->idx+p2->idx;
            if(sum) Attach(sum,p1->exp,&rear);
            p1 = p1->next;
            p2 = p2->next;
        }else if(p1->exp>p2->exp){
            Attach(p1->idx,p1->exp,&rear);
            p1 = p1->next;
        }else{
            Attach(p2->idx,p2->exp,&rear);
            p2 = p2->next;
        }
    }
    while(p1!=NULL){
        Attach(p1->idx,p1->exp,&rear);
        p1=p1->next;
    }
    while(p2!=NULL){
        Attach(p2->idx,p2->exp,&rear);
        p2=p2->next;
    }

    temp = PS;
    PS = PS->next;
    free(temp);
    return PS;
}


int main() {
    struct NODE* P1;
    struct NODE* P2;
    P1=build(),P2=build();
    mysotr(P1),mysotr(P2);

    struct NODE* P3;
    struct NODE* P4;
    P3= Add(P1,P2);
    printf("多项式相加的结果是：");
    myprint(P3);


    return 0;
}
//4 3 4 -5 2 6 1 -2 0
//3 5 20 -7 4 3 1

//4 -9 -1 2 2 3 3 2 1
//3 8 1 -5 4 5 18