题Two--Add Two Numbers

本题主要考察的是对链表的操作。在解题之前，先复习一下链表的头插法和尾插法的操作

不管采用哪种方法，首先应创建表头，目的是使第一个实际节点和后面的节点是等同的，不会因为删除、插入等操作区分开考虑。

头插法：不断的将新节点插入到表头后面。在头结点（为了操作方便，在单链表的第一个结点之前附加一个结点，称为头结点。头结点的数据域可以存储数据标题、表长等信息，也可以不存储任何信息，其指针域存储第一个结点的首地址）H之后插入数据，其特点是读入的数据顺序与线性表的逻辑顺序正好相反



struct ListNode {

    int val;

    ListNode *next;

    ListNode(int x) : val(x), next(NULL) {}

};

则头插法的过程为：

ListNode *head, *p;

head->next = NULL;

for(int i = 0; i < N; i++)

{

    p->next = head->next;

    head->next = p;

}

尾插法：将每次插入的新结点放在链表的尾部



则尾插法的过程为：

ListNode *head, *r, *s;

head->next = NULL;

r = head;

for(int i = 0; i < N; i++)

{

   s->next = r->next;

   r->next = s;

   r = s;

}

再来看本题，参考别人的博客分别使用头插法和尾插法解决

解法1：

#include<iostream>

#include<list>

using namespace std;

struct ListNode{

  int val;

  ListNode *next;

  ListNode(int x) : val(x), next(NULL){}

};

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)

{

   ListNode retHead(-1), *pr = &retHead, *p1 = NULL, *p2 = NULL;

   int carry = 0;

   for(p1 = l1, p2 = l2; p1 != NULL && p2 !=NULL; p1 = p1->next, p2 = p2->next)

   {

      int sum = p1->val + p2->val + carry;

      pr->next = new ListNode(sum % 10);

      pr = pr->next;

      carry = sum / 10;

    }

    for(ListNode *p = (p1 == NULL ? p2 : p1); p != NULL; p = p->next)

    {

      int sum = p->val + carry;

      pr->next = new ListNode(sum % 10);

      pr = pr->next;

      carry = sum / 10;

     }

     if(carry != 0)

     {

       pr->next = new ListNode(carry);

     }

     

     return retHead.next;

 }

 

 int main()

 {

   ListNode* a1 = new ListNode(9);

   ListNode* a2 = new ListNode(9);

   a1->next = a2;

   ListNode* b1 = new ListNode(1);

   ListNode* result = addTwoNumbers(a1, b1);

   while(result != NULL){

      cout << result->val << " ";

      result = result->next;

   }

   return
4000
0;

   

     

 }

解法2：

#include <iostream>

#include <stdio.h>

#include <stdlib.h>

using namespace std;

 

struct ListNode {

    int val;

    ListNode *next;

    ListNode(int x) : val(x), next(NULL) {}

};

 

class Solution {

public:

    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {

        ListNode *head = (ListNode *)malloc(sizeof(ListNode));

        ListNode *pre = head;

        ListNode *node = NULL;

      

        int c = 0,sum;

        while(l1 != NULL && l2 != NULL){

            sum = l1->val + l2->val + c;

            c = sum / 10;

            node = (ListNode *)malloc(sizeof(ListNode));

            node->val = sum % 10;

            node->next = NULL;

            //尾插法

            pre->next = node;

            pre = node;

            l1 = l1->next;

            l2 = l2->next;

        }

        //例如：2->4->3->1   5->6->4

        while(l1 != NULL){

            sum = l1->val + c;

            c = sum / 10;

            node = (ListNode *)malloc(sizeof(ListNode));

            node->val = sum % 10;

            node->next = NULL;

            //尾插法

            pre->next = node;

            pre = node;

            l1 = l1->next;

        }

        //例如：2->4->3   5->6->4->1

        while(l2 != NULL){

            sum = l2->val + c;

            c = sum / 10;

            node = (ListNode *)malloc(sizeof(ListNode));

            node->val = sum % 10;

            node->next = NULL;

            //尾插法

            pre->next = node;

            pre = node;

            l2 = l2->next;

        }

        //最后一位还有进位

        if(c > 0){

            node = (ListNode *)malloc(sizeof(ListNode));

            node->val = c;

            node->next = NULL;

            //尾插法

            pre->next = node;

            pre = node;

        }

        return head->next;

    }

};

int main() {

    Solution solution;

    int A[] = {2,4,7,9};

    int B[] = {5,6,4};

    ListNode *head = NULL;

    ListNode *head1 = (ListNode*)malloc(sizeof(ListNode));

    ListNode *head2 = (ListNode*)malloc(sizeof(ListNode));

    head1->next = NULL;

    head2->next = NULL;

    ListNode *node;

    ListNode *pre = head1;

    for(int i = 0;i < 4;i++){

        node = (ListNode*)malloc(sizeof(ListNode));

        node->val = A[i];

        node->next = NULL;

        pre->next = node;

        pre = node;

    }

    pre = head2;

    for(int i = 0;i < 3;i++){

        node = (ListNode*)malloc(sizeof(ListNode));

        node->val = B[i];

        node->next = NULL;

        pre->next = node;

        pre = node;

    }

    head = solution.addTwoNumbers(head1->next,head2->next);

    while(head != NULL){

        printf("%d ",head->val);

        head = head->next;

    }

    return 0;

}