链表系列之单链表——使用单链表实现大整数相加

来源：互联网发布：net是什么域名编辑：程序博客网时间：2024/05/18 03:11

大数相加在我之前的一篇博客里有一个使用数组实现的方案，使用单链表实现更灵活。

有两个由单链表表示的数。每个结点代表其中的一位数字。

数字的存储是逆序的，也就是说个位位于链表的表头。
写一函数使这两个数相加并返回结果，结果也由链表表示。

eg.

Input:(3->9->6), (4->7->8->3)

Output:(7->6->5->4)

#include <iostream>#include <stack>using namespace std;typedef struct node{int data;node *next;}Node, *LinkList;//建立链表Node* createList(const int a[], int n){Node *head, *endPtr;head = endPtr = NULL;for(int i=0;i<n;i++){Node *temp = new Node;temp->data = a[i];temp->next = NULL;if(i==0){head = endPtr = temp;}else{endPtr->next = temp;endPtr = temp;}}return head;}/*从尾到头打印链表，要求不修改链表结构*///使用栈适配器void PrintListReversing(LinkList pHead){stack<Node*> nodes;Node* pNode = pHead;if(pNode==NULL)return;while(pNode!=NULL)   //将节点依次入栈{nodes.push(pNode);pNode = pNode->next;}while(!nodes.empty())   //出栈{pNode = nodes.top();   //读取栈顶元素cout<<pNode->data;nodes.pop();   //删除栈顶元素}}//大数相加Node *ListAdd(Node* L1, Node* L2){if(L1==NULL)return L2;if(L2==NULL)return L1;Node *ptr1 = L1, *ptr2 = L2, *ResultPtr=NULL, *TmpPtr=NULL;int carry = 0;Node *p_node = new Node();p_node->data = (L1->data+L2->data)%10;p_node->next = NULL;carry = (L1->data+L2->data)/10;ResultPtr = TmpPtr = p_node;TmpPtr->next = NULL;L1 = L1->next;L2 = L2->next;while(L1 && L2){Node *pNode = new Node();TmpPtr->next = pNode;int tmp = L1->data+L2->data+carry;carry = tmp/10;pNode->data = tmp%10;pNode->next = NULL;TmpPtr = TmpPtr->next;L1 = L1->next;L2 = L2->next;}while(L1){Node *pNode = new Node();TmpPtr->next = pNode;int tmp = L1->data+carry;carry = tmp/10;pNode->data = tmp%10;pNode->next = NULL;TmpPtr = TmpPtr->next;L1 = L1->next;}while(L2){Node *pNode = new Node();TmpPtr->next = pNode;int tmp = L2->data+carry;carry = tmp/10;pNode->data = tmp%10;pNode->next = NULL;TmpPtr = TmpPtr->next;L2 = L2->next;}if(carry){Node *pNode = new Node();TmpPtr->next = pNode;pNode->data = carry;pNode->next = NULL;}return ResultPtr;}int main(){int a[] = {1,9,9};  //991int b[] = {9,8,5,6,6,2,8};   //8266589Node *L1 = createList(a,3), *L2 = createList(b,7), *L3 = NULL;L3 = ListAdd(L1,L2);PrintListReversing(L1);cout<<"+";PrintListReversing(L2);cout<<"=";PrintListReversing(L3);cout<<endl;return 1;}

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