已知两个稀疏一元多项式,通过线性表的链式存储结构实现两个一元多项式的求和,要求计算的过程中空间复杂度为O(n)输入:A(x)=7+3x+9x8+5x17;B=8x+22x7-9x8输出:C(x)=7+11x+22x7+5x17用C语言代码实现
时间: 2024-11-18 09:17:04 浏览: 9
在C语言中,我们可以使用动态内存分配和链式数据结构来实现两个稀疏一元多项式的求和,每个节点表示多项式的系数和对应的指数。由于只需要存储非零项,所以空间复杂度可以达到O(n),其中n是两个多项式中非零项的总数。
首先,我们需要定义一个结构体`Term`来表示单个项:
```c
typedef struct {
int coefficient; // 系数
int exponent; // 指数
} Term;
```
然后我们创建链表结构来存储多项式:
```c
typedef struct Polynomial {
Term* terms; // 链表头指针
int degree; // 最大指数
} Polynomial;
```
接下来编写函数来合并两个多项式并返回结果:
```c
void addPolynomial(Polynomial* result, const Polynomial* a, const Polynomial* b) {
if (a->degree < b->degree) {
Polynomial temp = *a;
a = b;
b = &temp;
}
Term* newTerm = NULL;
for (Term* current = a->terms; current != NULL; current = current->next) {
int combinedCoefficient = current->coefficient + (current->exponent == b->degree ? b->terms->coefficient : 0);
if (combinedCoefficient != 0 || newTerm == NULL) {
if (newTerm == NULL) {
newTerm = malloc(sizeof(Term));
newTerm->coefficient = combinedCoefficient;
newTerm->exponent = current->exponent;
} else {
Term* newNode = malloc(sizeof(Term));
newNode->coefficient = combinedCoefficient;
newNode->exponent = current->exponent;
newTerm->next = newNode;
}
if (current->exponent > result->degree)
result->degree = current->exponent;
}
}
// 添加B剩余的项(如果有的话)
while (b->terms != NULL) {
if (result->degree >= b->terms->exponent) {
Term* newNode = malloc(sizeof(Term));
newNode->coefficient = b->terms->coefficient;
newNode->exponent = b->terms->exponent;
result->degree = b->terms->exponent;
newTerm->next = newNode;
}
b->terms = b->terms->next;
}
result->terms = newTerm;
}
// 使用示例
int main() {
Polynomial A = { .terms = NULL, .degree = 0 };
Polynomial B = { .terms = NULL, .degree = 0 };
// 初始化多项式A和B...
// ...
Polynomial C = { .terms = NULL, .degree = 0 };
addPolynomial(&C, &A, &B);
return 0;
}
```
注意:这个实现假设了多项式初始化时已经完成了对系数和指数的设置,并且`malloc`操作成功。在实际应用中,需要处理可能出现的内存分配失败的情况。
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