2-1 Add Two Polynomials (20 分)
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2-1 Add Two Polynomials (20 分)
Write a function to add two polynomials. Do not destroy the input. Use a linked list implementation with a dummy head node. Note: The zero polynomial is represented by an empty list with only the dummy head node.
Format of functions:
Polynomial Add( Polynomial a, Polynomial b );where Polynomial is defined as the following:
typedef struct Node *PtrToNode;
struct Node {
int Coefficient;
int Exponent;
PtrToNode Next;
};
typedef PtrToNode Polynomial;
/* Nodes are sorted in decreasing order of exponents.*/ The function Add is supposed to return a polynomial which is the sum of a and b.
Sample program of judge:
#include <stdio.h>
#include <stdlib.h>
typedef struct Node *PtrToNode;
struct Node {
int Coefficient;
int Exponent;
PtrToNode Next;
};
typedef PtrToNode Polynomial;
Polynomial Read(); /* details omitted */
void Print( Polynomial p ); /* details omitted */
Polynomial Add( Polynomial a, Polynomial b );
int main()
{
Polynomial a, b, s;
a = Read();
b = Read();
s = Add(a, b);
Print(s);
return 0;
}
/* Your function will be put here */Sample Input:
4
3 4 -5 2 6 1 -2 0
3
5 20 -7 4 3 1Sample Output:
5 20 -4 4 -5 2 9 1 -2 01.系数为0要跳过
2.上来给你个空表，反手就把另一个返回去
3.记住链表头结点是空的，第二个才存数据
Polynomial Add( Polynomial a, Polynomial b ){
if(a == NULL) return b;
if(b == NULL) return a;
Polynomial ans, tmp, before; //ans是头结点, before指向tmp前一个结点
before = (Polynomial)malloc(sizeof(Polynomial));
ans = before;
a = a->Next;
b = b->Next;
while(a != NULL && b != NULL){
tmp = (Polynomial)malloc(sizeof(Polynomial));
before->Next = tmp;
if(a->Exponent == b->Exponent){
tmp->Exponent = a->Exponent;
tmp->Coefficient = a->Coefficient + b->Coefficient;
a = a->Next;
b = b->Next;
if(tmp->Coefficient == 0)continue;//0不存
}
else if(a->Exponent > b->Exponent){
tmp->Exponent = a->Exponent;
tmp->Coefficient = a->Coefficient;
a = a->Next;
}
else{
tmp->Exponent = b->Exponent;
tmp->Coefficient = b->Coefficient;
b = b->Next;
}
before = tmp;
}
if(a != NULL) before->Next = a;//后面全接上剩下的部分
else before->Next = b;//
return ans;
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原始发表：2019-09-22 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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