Add Two Numbers

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

注意:最后一位如果有进位,需要新创建一个结点来存放。

C++代码实现:

#include<iostream>
#include<new>
using namespace std;

//Definition for singly-linked list.
struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};
class Solution
{
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
    {
        ListNode *p=l1;
        ListNode *pre=l1;
        ListNode *q=l2;
        int carry=0;
        while(p&&q)
        {
            if(p->val+q->val+carry>9)
            {
                p->val=(p->val+q->val+carry)%10;
                carry=1;
            }
            else
            {
                 p->val+=q->val+carry;
                 carry=0;
            }
            pre=p;
            p=p->next;
            q=q->next;
        }
        if(q)
            pre->next=q;
        while(q)
        {
            if(q->val+carry>9)
            {
                q->val=(q->val+carry)%10;
                carry=1;
            }
            else
            {
                q->val+=carry;
                carry=0;
                break;
            }
            pre=q;
            q=q->next;
        }
        while(p)
        {
            if(p->val+carry>9)
            {
                p->val=(p->val+carry)%10;
                carry=1;
            }
            else
            {
                p->val+=carry;
                carry=0;
                break;
            }
            pre=p;
            p=p->next;
        }
        if(carry)
        {
            p=new ListNode(1);
            pre->next=p;
        }
        return l1;
    }
    void createList(ListNode *&head,int *arr)
    {
        ListNode *p=NULL;
        int i=0;
        for(i=0; i<5; i++)
        {
            if(head==NULL)
            {
                head=new ListNode(arr[i]);
                if(head==NULL)
                    return;
            }
            else
            {
                p=new ListNode(arr[i]);
                p->next=head;
                head=p;
            }
        }
    }
};

int main()
{
    Solution s;
    ListNode *L1=NULL;
    ListNode *L2=NULL;
    ListNode *L=NULL;
    int arr1[10]= {1,9,7,5,3};
    int arr2[10]= {0,8,6,4,2};
    s.createList(L1,arr1);
    s.createList(L2,arr2);
    L=s.addTwoNumbers(L1,L2);
    while(L)
    {
        cout<<L->val<<" ";
        L=L->next;
    }
}

运行结果:

#include<iostream>
#include<new>
using namespace std;

//Definition for singly-linked list.
struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution
{
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
    {
        int carry=0;
        ListNode *pre=l1;
        ListNode *p=l1;
        ListNode *q=l2;
        while(p&&q)
        {
            p->val+=q->val+carry;
            cout<<p->val<<endl;
            if(p->val>9)
            {
                carry=1;
                p->val=p->val%10;
            }
            else
                carry=0;
            pre=p;
            p=p->next;
            q=q->next;
        }
        if(q)
        {
            pre->next=q;
            p=q;
        }
        while(p)
        {
            p->val=p->val+carry;
            if(p->val>9)
            {
                p->val=p->val%10;
                carry=1;
            }
            else
            {
                carry=0;
                break;
            }
            pre=p;
            p=p->next;
        }
        if(carry==1)
            pre->next=new ListNode(1);
        return l1;
    }
    void createList(ListNode *&L,int arr[],int n)
    {
        int i;
        for(i=0; i<n; i++)
        {
            ListNode *tmp=new ListNode(arr[i]);
            if(L==NULL)
            {
                L=tmp;
            }
            else
            {
                tmp->next=L;
                L=tmp;
            }
        }
    }
};

int main()
{
    Solution s;
    ListNode *L1=NULL;
    ListNode *L2=NULL;
    ListNode *L=NULL;
    int arr1[10]= {8,9};
    int arr2[10]= {1};
    s.createList(L1,arr1,2);
    s.createList(L2,arr2,1);
    L=s.addTwoNumbers(L1,L2);
    while(L)
    {
        cout<<L->val<<" ";
        L=L->next;
    }
}

  

原文地址:https://www.cnblogs.com/wuchanming/p/4100195.html