例如:9->9->9->NULL
+ 1->NULL
1->0->0->0->NULL
思路:
使用递归,能够实现从前往后计算。
// LinkTable.cpp : 定义控制台应用程序的入口点。
//
#include "stdafx.h"
#include <iostream>
#include <string>
using namespace std;
//链表的结构体
struct node
{
char val;
node * next;
};
//建立链表
struct node * create( string & str_link )
{
int len = str_link.length();
struct node * phead = new node(); //带有表头的链表,表头中不存储任何元素
struct node * preNode = phead;
for( int i=0; i<len; i++ )
{
struct node * pNode = new node();
pNode->val = str_link[i];
pNode->next = NULL;
preNode->next = pNode;
preNode = pNode;
}
return phead;
}
//输出链表
void out_link( struct node * phead )
{
if( phead == NULL )
return;
struct node * pNode = phead->next;
while( pNode )
{
cout <<pNode->val;
pNode = pNode->next;
}
cout << endl;
}
//求无表头链表的长度
//返回-1为链表不存在
int link_length( struct node* pNode )
{
if(!pNode) return -1;
int len=0;
while( pNode )
{
pNode = pNode->next;
len++;
}
return len;
}
//大数相加递归算法
//pNode1, pNode2为两个中间运算结点,但不是头结点
struct node * add( struct node * pNode1, struct node * pNode2, int & carry )
{
if( !pNode1 ) return pNode2;
if( !pNode2 ) return pNode1;
//为了参数简洁,这里增大了计算量,可以将链表长度作为参数传进来
int len1 = link_length( pNode1 );
int len2 = link_length( pNode2 );
if( len1 == len2 )
{
if( len1==1 ) //递归终止条件
{
struct node * pNode = new node();
int sum = (pNode1->val - '0' ) + ( pNode2->val -'0');
carry = sum/10;
pNode->val = sum%10 + '0';
pNode->next = NULL;
return pNode;
}
else
{
int carry_cur=0;
struct node * pNode = new node();
struct node * pNext = add( pNode1->next, pNode2->next, carry_cur );
int sum = (pNode1->val - '0' ) + ( pNode2->val -'0') + carry_cur;
carry = sum/10;
pNode->val = sum%10 + '0';
pNode->next = pNext;
return pNode;
}
}
if( len1>len2 )
{
int carry_cur=0;
struct node * pNode = new node();
struct node * pNext = add( pNode1->next, pNode2, carry_cur );
int sum = (pNode1->val - '0' ) + carry_cur;
carry = sum/10;
pNode->val = sum%10 + '0';
pNode->next = pNext;
return pNode;
}
if( len1<len2 )
{
int carry_cur=0;
struct node * pNode = new node();
struct node * pNext = add( pNode1, pNode2->next, carry_cur );
int sum = (pNode2->val - '0' ) + carry_cur;
carry = sum/10;
pNode->val = sum%10 + '0';
pNode->next = pNext;
return pNode;
}
return NULL;
}
struct node * add( struct node * phead1, struct node * phead2 )
{
if( !phead1 || !phead1->next ) return phead2;
if( !phead2 || !phead2->next ) return phead1;
int carry = 0;
struct node * pNode = add( phead1->next, phead2->next, carry );
if( carry > 0 ) //有进位,则需要多一个结点
{
struct node * pCarry = new node();
pCarry->val = '0' + carry;
pCarry->next = pNode;
pNode = pCarry;
}
struct node * phead = new node();
phead->next = pNode;
return phead;
}
void test()
{
string str;
cout << "Input the first link:"<<endl;
cin >> str;
struct node *phead1 = create( str );
cout << "Input the second link:"<<endl;
cin >> str;
struct node *phead2 = create( str );
struct node * phead = add( phead1, phead2);
cout<< "The result is:" <<endl;
out_link( phead );
}
int _tmain(int argc, _TCHAR* argv[])
{
test();
return 0;
}