链表相加

链表相加：

给定两个链表，分别表示两个非负整数。它们的数字逆序存储在链表中，且每个结点中存储一个数字，计算两个数的和，并且返回和的链表头指针。

如：输入：2->4->3，5->6->4.

输出：7->0->8.

程序实现：

  1 /************************************
  2 File Name:ListAdd.cpp
  3 Author: godfrey
  4 Created Time: 2016/04/28
  5 *************************************/
  6 #include <iostream>
  7 #include <cstdio>
  8 #include <cstdlib>
  9 using namespace std;
 10 
 11 typedef struct tagSNode{
 12     int value;
 13     tagSNode* pNext;
 14 
 15     tagSNode(int v):value(v),pNext(NULL) {}
 16 }SNode;
 17 //打印链表
 18 void Print(SNode* pHead){
 19     SNode* p = pHead->pNext;
 20     while(p){
 21         cout<<p->value<<" ";
 22         p = p->pNext;
 23     }
 24     cout<<endl;
 25 }
 26 //删除分配结点空间
 27 void Destroy(SNode* pHead){
 28     SNode* p = pHead->pNext;
 29     while(p){
 30         pHead->pNext = p->pNext;//与头插法相反
 31         delete p;
 32         p = pHead->pNext;
 33     }
 34     delete pHead;
 35 }
 36 //链表相加
 37 SNode* ListAdd(SNode* pHead1,SNode* pHead2){
 38     SNode* pSum = new SNode(0);
 39     SNode* pTail = pSum;//新结点插入pTail的后面
 40     SNode* p1 = pHead1->pNext;
 41     SNode* p2 = pHead2->pNext;
 42     SNode* pCur = NULL;
 43     int carry = 0;//进位
 44     int value;
 45     //处理两者都存在的加法
 46     while(p1&&p2){
 47         value = p1->value + p2->value +carry;
 48         carry = value / 10;
 49         value %= 10;
 50         pCur = new SNode(value);
 51         pTail->pNext = pCur;
 52         pTail = pCur;
 53 
 54         p1 = p1->pNext;
 55         p2 = p2->pNext;
 56     }
 57     //处理长的链表
 58     SNode* p = p1 ? p1 : p2;
 59     while(p){
 60         value = p->value +carry;
 61         carry = value / 10;
 62         value %= 10;
 63         pCur = new SNode(value);
 64         pTail->pNext = pCur;
 65         pTail = pCur;
 66 
 67         p = p->pNext;
 68     }
 69     //处理可能存在的进位
 70     if(carry!=0){
 71         pTail->pNext = new SNode(carry);
 72     }
 73 
 74     return pSum;
 75 }
 76 
 77 int main()
 78 {
 79     SNode* pHead1 = new SNode(0);
 80     for(int i=0;i<6;i++){
 81         SNode* p = new SNode(rand()%10);
 82         p->pNext = pHead1->pNext;
 83         pHead1->pNext = p;
 84     }
 85 
 86     SNode* pHead2 = new SNode(0);
 87     for(int i=0;i<9;i++){
 88         SNode* p = new SNode(rand()%10);
 89         p->pNext = pHead2->pNext;
 90         pHead2->pNext = p;
 91     }
 92     Print(pHead1);
 93     Print(pHead2);
 94     SNode* pSum = ListAdd(pHead1,pHead2);
 95     Print(pSum);
 96     Destroy(pHead1);
 97     Destroy(pHead2);
 98     Destroy(pSum);
 99     return 0;
100 }

运行结果：

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posted on 2016-04-28 14:51  Brainer-Gao  阅读(230)  评论(0)  编辑  收藏  举报

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