程序员面试金典： 9.2链表 2.5对两个用链表表示的整数求和

#include <iostream>#include <stdio.h>using namespace std;const int MAXSIZE = 1000;/*问题：给定两个用链表表示的整数，每个节点包含一个数位。这些数位是反向存放的，也就是      个位排在链表首部。编写函数对这两个整数求和，并用链表形式返回结果。分析：比如两个数分别是123,789,由于个位在首部，得到如下的两个数      3 2 1  9 8 7  2 1 9  最终结果为912，算的时候直接从左向右（从低位到高位算即可，读取的时候反向读取链表即可）输入:123 789123 89输出:912212关键:1 需要明白相加的两个整数长度可能不同，需要注意遍历*/typedef struct Node{int value;Node* pNext;}Node;//构建连边，按照尾插法来做，返回节点值到出现次数的映射void buildList(int* pArray , Node* head , int num){if(pArray == NULL){return;}if(head == NULL){return;}//尾插法： 保留最后一个结尾节点，将新生成的节点插入在结尾节点，并令结尾节点为当前节点Node* pLast = head;//int num = sizeof(pArray) / sizeof(int);for(int i = 0 ; i < num ; i++){int value = *(pArray + i);Node* pNode = new Node();pNode->value = value;pLast->pNext = pNode;pLast = pNode;}}//这里必须先打印链表末尾，再打印末尾前面，void printList(Node* pHead){if(NULL == pHead){return;}printList(pHead->pNext);cout << pHead->value;}void releaseList(Node* pHead){if(NULL == pHead){return;}Node* pNode = pHead->pNext;Node* pPrevious = pHead;while(pNode){Node* pDeleteNode = pNode;pPrevious->pNext = pNode->pNext;pNode = pNode->pNext;pPrevious = pPrevious->pNext;delete pDeleteNode;}//删除头结点delete pHead;}//将123除10取余，123 / 10 =12余3 ， 12 / 10 = 1余2 ， 1 / 10 = 0余1 ，依次得到个位，10位，百位//获取总共的位数int getDigitNum(int num){int temp = num;int count = 0;do{temp /= 10;count++;}while(temp != 0);return count;}void getDigitArr(int num,int* pArr){int temp = num;int count = 0;//这里先拿到个位的，个位需要放在链表前面do{int num = temp % 10;pArr[count] = num;temp /= 10;count++;}while(temp != 0);}//将两个链表求和,个位是在链表首位，所加结果放在第一个链表中并返回,需要放在较长的链表中(默认是第一个)Node* sum(Node* nNode,Node* mNode){if(nNode == NULL || mNode == NULL){return NULL;}Node* pNNode = nNode->pNext;Node* pMNode = mNode->pNext;int pass = 0; //设置进位//注意，这里只能有一个不为空即可，如果另一个数为空，则按0计算while(pNNode || pMNode){//所加结果需要进位if(pMNode == NULL){pNNode->value += 0 + pass;}else{pNNode->value += pMNode->value + pass;}if( pNNode->value > 10){pNNode->value -= 10;pass = 1;}else{pass = 0;}pNNode = pNNode->pNext;if(pMNode != NULL){pMNode = pMNode->pNext;}}return nNode;}int main(int argc, char* argv[]){int n,m;while(cin >> n >> m){int nDigitNum = getDigitNum(n);int mDigitNum = getDigitNum(m);int* nArr = new int[nDigitNum];int* mArr = new int[mDigitNum];getDigitArr(n, nArr);getDigitArr(m, mArr);//得到两个数组之后，下面需要存入到链表中Node* nNode = new Node();buildList(nArr , nNode , nDigitNum);Node* mNode = new Node();buildList(mArr , mNode , mDigitNum);Node* pResult;if(nDigitNum >= mDigitNum){pResult = sum(nNode,  mNode);}else{pResult = sum(mNode,  nNode);}//注意这里空节点需要排除printList(pResult->pNext);cout << endl;releaseList(nNode);//releaseList(nNode);releaseList(mNode);delete[] nArr;delete[] mArr;}system("pause");return 0;}