该代码的原理：首先创建两个链表tempList1和tempList2分别存放按照指数从小到大一定长度的数据。再申请五个指针p,q,r,s,temp指针p和q分别指向tempList1（tempList2)和tempList2(tempList1),s是free指数相同的结点，r将两个链表的结点相连以及连接free(s)结点的前后两个结点，按照我的代码首先p指向tempList1的第一个结点（头结点的下一个结点）q指向tempList2的第一个结点（头结点的下一个结点），然后进行结点所存储的指数的比较，分三种情况，第一：p的指数小于q的指数，r = p,p = p->next,然后继续比较p和q的指数的大小，第二：p的指数大于q的指数，这一种情况相当重要，将p的前一个结点与q连接，这个时候一定要注意一个细节此时的tempList1已经不再与p相连，相当于tempList1已经连接了tempList2,如果要进行下一步操作一定要将p和q交换位置，第三：p的指数和q的指数相等时，情况一：当系数相加不等于0时，p的指数直接等于p,q指数相加,然后s = q,q = q->next,free(s)。情况二：当系数相加等于0时，p,q都指向下一个结点，然后free两个指数相同的结点，此时问题来了，如果free了tempList1的结点，那么整个链表就没有连起来了，此时就要r->next = p,这样就可以重新将链表连起来，如果没有连有可能在打印链表的时候就会形成死循环，因为整个链表没有指向空的结点。最后在将提前结束的结点的NULL指向没有结束的结点。这样就完成了多项式的加法。
上代码#include#include typedef struct LinkNode{ int coefficient; int exponent; struct LinkNode *next; } *LinkList, *NodePtr; LinkList initLinkList(){ LinkList tempHeader = (LinkList)malloc(sizeof(struct LinkNode)); tempHeader->coefficient = 0; tempHeader->exponent = 0; tempHeader->next = NULL; return tempHeader; }// Of initLinkList void printList(LinkList paraHeader){ NodePtr p = paraHeader->next; while (p->next != NULL) { printf("%d * 10^%d + ", p->coefficient, p->exponent); p = p->next; }// Of while printf("%d * 10^%d",p->coefficient,p->exponent); printf("rn"); }// Of printList void printNode(NodePtr paraPtr, char paraChar){ if (paraPtr == NULL) { printf("NULLrn"); } else { printf("The element of %c is (%d * 10^%d)rn", paraChar, paraPtr->coefficient, paraPtr->exponent); }// Of while }// Of printNode void appendElement(LinkList paraHeader, int paraCoefficient, int paraExponent){ NodePtr p, q; // Step 1. Construct a new node. q = (NodePtr)malloc(sizeof(struct LinkNode)); q->coefficient = paraCoefficient; q->exponent = paraExponent; q->next = NULL; // Step 2. Search to the tail. p = paraHeader; while (p->next != NULL) { p = p->next; }// Of while // Step 3. Now add/link. p->next = q; }// Of appendElement void add(NodePtr paraList1, NodePtr paraList2){ NodePtr p, q, r, s,temp; //p,q分别指向tempList1和tempList2 //r作用一：将tempList1和tempList2的结点相连,作用二：将删除的结点的前后结点相连 //s删除指定结点 //temp将p,q指向交换 // Step 1. Search to the position. p = paraList1->next; printNode(p, 'p'); q = paraList2->next; printNode(q, 'q'); r = paraList1; // Previous pointer for inserting. printNode(r, 'r'); free(paraList2); // The second list is destroyed. while ((p != NULL) && (q != NULL)) { if (p->exponent < q->exponent) { //Link the current node of the first list. printf("case 1rn"); r = p; printNode(r, 'r'); p = p->next; printNode(p, 'p'); } else if ((p->exponent > q->exponent)) { //Link the current node of the second list. printf("case 2rn"); r->next = q; r = q; printNode(r, 'r'); q = q->next; printNode(q, 'q'); temp = p;//交换q,p的指向 p = q; q = temp; } else { printf("case 3rn"); //Change the current node of the first list. p->coefficient = p->coefficient + q->coefficient; printf("The coefficient is: %d.rn", p->coefficient); if (p->coefficient == 0) { printf("case 3.1rn"); s = p; p = p->next; printNode(p, 'p'); free(s); r->next = p;//这一步非常重要 } else { printf("case 3.2rn"); r = p; printNode(r, 'r'); p = p->next; printNode(p, 'p'); }// Of if s = q; q = q->next; //printf("q is pointing to (%d, %d)rn", q->coefficient, q->exponent); free(s); }// Of if printf("p = %ld, q = %ld rn", p, q); } // Of while printf("End of while.rn"); if (p == NULL) { r->next = q; } else { r->next = p; } // Of if printf("Addition ends.rn"); }// Of add void additionTest(){ // Step 1. Initialize the first polynomial. LinkList tempList1 = initLinkList(); appendElement(tempList1, 7, 2); appendElement(tempList1, 3, 3); appendElement(tempList1, 9, 8); appendElement(tempList1, 5, 9); appendElement(tempList1, 4, 11); appendElement(tempList1, 1, 20); printList(tempList1); // Step 2. Initialize the second polynomial. LinkList tempList2 = initLinkList(); appendElement(tempList2, 8, 1); appendElement(tempList2, 22, 7); appendElement(tempList2, -9, 8); appendElement(tempList2, 6, 11); printList(tempList2); // Step 3. Add them to the first. add(tempList1, tempList2); printList(tempList1); }// Of additionTest void main(){ additionTest(); printf("Finish.rn"); }// Of main
运行结果
7 * 10^2 + 3 * 10^3 + 9 * 10^8 + 5 * 10^9 + 4 * 10^11 + 1 * 10^20 8 * 10^1 + 22 * 10^7 + -9 * 10^8 + 6 * 10^11 The element of p is (7 * 10^2) The element of q is (8 * 10^1) The element of r is (0 * 10^0) case 2 The element of r is (8 * 10^1) The element of q is (22 * 10^7) p = 2037088, q = 2036832 case 2 The element of r is (7 * 10^2) The element of q is (3 * 10^3) p = 2036864, q = 2037088 case 1 The element of r is (3 * 10^3) The element of p is (9 * 10^8) p = 2036896, q = 2037088 case 2 The element of r is (22 * 10^7) The element of q is (-9 * 10^8) p = 2037120, q = 2036896 case 3 The coefficient is: 0. case 3.1 The element of p is (6 * 10^11) p = 2037152, q = 2036928 case 2 The element of r is (5 * 10^9) The element of q is (4 * 10^11) p = 2036960, q = 2037152 case 3 The coefficient is: 10. case 3.2 The element of r is (10 * 10^11) The element of p is (1 * 10^20) p = 2036992, q = 0 End of while. Addition ends. 8 * 10^1 + 7 * 10^2 + 3 * 10^3 + 22 * 10^7 + 5 * 10^9 + 10 * 10^11 + 1 * 10^20 Finish.方法二
该方法是将tempList1的每一个结点的指数都与tempList2里面的小于该指数的结点进行比较，如果在tempList2里面找到了相同指数的结点，那么就直接将其free掉，如果系数相加等于零也将tempList1里面的结点也free掉，如果相加系数不等于零则将tempList1里面的结点保留，系数赋值为相加的系数，然后将tempList1的尾结点与tempList2的第一个结点（头结点的下一个结点）相连，最后将tempList1按照指数从小到大的顺序排列。
上代码#include#include struct Node { int coefficient; int exponent; struct Node* next; }; struct Node* createList() { struct Node* headNode = (struct Node*)malloc(sizeof(struct Node)); headNode->coefficient = 0; headNode->exponent = 0; headNode->next = NULL; return headNode; } struct Node* createNode(int coefficient,int exponent) { struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); newNode->coefficient = coefficient; newNode->exponent = exponent; newNode->next = NULL; return newNode; } void printList(struct Node* headNode) { struct Node* pMove = headNode->next; while(pMove->next != NULL) { printf(" %d * 10^%d +",pMove->coefficient,pMove->exponent); pMove = pMove->next; } printf(" %d * 10^%drn",pMove->coefficient,pMove->exponent); } void insertNodeBytail(struct Node* headNode,int coefficient,int exponent) { struct Node* newNode = createNode(coefficient,exponent); struct Node* p = headNode; while(p->next != NULL) { p = p->next; } p->next = newNode; } //将tempList1的结点按照指数从小到大的顺序排列 void sortList(struct Node* headNode) { struct Node* p = NULL; struct Node* q = NULL; struct Node* tempNode = createList(); for(p = headNode->next; p != NULL; p = p->next) { for(q = p->next; q != NULL; q = q->next) { if(p->exponent > q->exponent) { tempNode->coefficient = p->coefficient; tempNode->exponent = p->exponent; p->coefficient = q->coefficient; p->exponent = q->exponent; q->coefficient = tempNode->coefficient; q->exponent = tempNode->exponent; } } } } //两个链表的多项式相加 void addList(struct Node* tempList1,struct Node* tempList2) { struct Node* p = tempList1->next; struct Node* q = tempList2->next; struct Node* r = tempList2;//作用：当tempList2的结点的指数等于templist1的指数时删除tempList2结点 struct Node* s;//充当删除结点 struct Node* temp;//当结点相加的系数等于0时临时储存p的地址防止在删除templist1的结点时误删本来系数就等于0的结点 int tempList1_exponent;//存储tempList1的指数 int tempList2_exponent;//存储tempList2的指数 while(p != NULL) { tempList1_exponent = p->exponent; tempList2_exponent = tempList2->next->exponent; while(tempList2_exponent <= tempList1_exponent) { if(tempList1_exponent != tempList2_exponent)//不等就往tempList2的下一个结点走 { r = q; q = q->next; if(q == NULL)//当q == NULL时q->exponent值没有程序就此终止所以这一步非常重要 { break; } tempList2_exponent = q->exponent; }else { p->coefficient = p->coefficient + q->coefficient; temp = p; s = q; q = q->next; r->next = q; free(s); if(q == NULL) { break; } tempList2_exponent = q->exponent; } } //s = p; p = p->next; //tempList1往下一个结点走 if(temp->coefficient == 0)//如果相加等于零就要删除在templist1里面的指数为tempList1_exponent的结点 { struct Node* m = tempList1; struct Node* n; while(m->next != NULL && m->next->exponent != tempList1_exponent) { m = m->next; } n = m->next; m->next = m->next->next; temp->coefficient = 1;//防止下一次循环误删必须赋值非零数 free(n); } q = tempList2->next; } struct Node* pMove = tempList1; while(pMove->next != NULL) { pMove = pMove->next; } pMove->next = tempList2->next; sortList(tempList1); free(tempList2); } void test() { struct Node* tempList1 = createList(); insertNodeBytail(tempList1, -8, 1); insertNodeBytail(tempList1, 7 ,2); insertNodeBytail(tempList1, 3, 3); insertNodeBytail(tempList1, 9, 8); insertNodeBytail(tempList1, 5, 9); insertNodeBytail(tempList1, 0, 20); printList(tempList1); struct Node* tempList2 = createList(); insertNodeBytail(tempList2, 8, 1); insertNodeBytail(tempList2, 22, 7); insertNodeBytail(tempList2, -9, 8); insertNodeBytail(tempList2, 6, 9); insertNodeBytail(tempList2, 0, 12); printList(tempList2); addList(tempList1,tempList2); printList(tempList1); } int main() { test(); return 0; }
运行结果
-8 * 10^1 + 7 * 10^2 + 3 * 10^3 + 9 * 10^8 + 5 * 10^9 + 0 * 10^20 8 * 10^1 + 22 * 10^7 + -9 * 10^8 + 6 * 10^9 + 0 * 10^12 7 * 10^2 + 3 * 10^3 + 22 * 10^7 + 11 * 10^9 + 0 * 10^12 + 0 * 10^20
