LeetCode 2 Add Two Sum 解题报告

LeetCode 2 Add Two Sum 解题报告

LeetCode第二题 Add Two Sum 首先我们看题目要求：

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

题目分析

这道题目是一道基础的链表题，给定两个非负数，它们是按照逆序存储的，每个节点值保留一个数值，要求输出这两个数之和，返回结果链表。本道题目主要是考察链表遍历的一些操作。

思路

1.首先用两个指针，分别同时遍历两个链表，按位相加，设置相应进位标志。
2.当两个链表长度不一致时，结束按位相加的遍历之后，将剩余链接接上
3.需要注意连续进位。

以下给出完整的测试代码，在这里为了操作方便，在遍历时统一把数据放到了list的容器中，主要是担心，对于链接，连续进位时直接用指针new出错，直接将每一位Push到list中，最后直接通过list构造出一个ListNode的链表

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

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

ListNode * createListNode( int * arr, int num)
{
    int i = 0;
    ListNode * head = new ListNode(arr[0]);//head pointer
    ListNode * p1 = head;
    ListNode * p2 = head;
    if(num == 1)
    {
        head->next = NULL;
        return head;
    }
    else
    {
        for(i = 1; i < num; i++)
        {
            p1 = new ListNode(arr[i]);
            p2->next = p1;
            p2 = p1;
        }
        p1->next = NULL;
    }
    return head;
}

class Solution
{
public:
     ListNode * createListNode2( list<int> iList)//
        {
        int num = iList.size();
        list<int>::iterator it = iList.begin();
        ListNode * head = new ListNode(*it);//head pointer
        ListNode * p1 = head;
        ListNode * p2 = head;
        it++;
        if(num == 1)
        {
            head->next = NULL;
            return head;
        }
        else
        {
            for(; it != iList.end(); it++)
            {
                p1 = new ListNode(*it);
                p2->next = p1;
                p2 = p1;
            }
            p1->next = NULL;
        }
        return head;
        }

        ListNode * addTwoNumbers (ListNode * ln1,ListNode * ln2)
        {

                list<int> result;
                ListNode * p;
                ListNode * p1 = ln1;
                ListNode * p2 = ln2;
                int carryFlag = 0;
                int curNum = 0;
                while(p1 != NULL && p2 != NULL)
                {
                    curNum  = (p1->val + p2->val + carryFlag)%10;
                    if((p1->val + p2->val + carryFlag) >= 10)
                        carryFlag = 1;
                    else
                        carryFlag = 0;
                    result.push_back(curNum);
                    p1 = p1->next;
                    p2 = p2->next;
                }
                if(p1 == NULL && p2 == NULL)
                {
                    if (carryFlag == 1)
                        result.push_back(carryFlag);
                }
                else if(p1 != NULL && p2 == NULL )
                {
                    while(p1 != NULL)
                    {
                        curNum = (p1->val+carryFlag) %10;
                        if(p1->val + carryFlag >= 10)
                            carryFlag = 1;
                        else
                            carryFlag = 0;
                        result.push_back(curNum);
                        p1 = p1->next;
                    }
                    if(carryFlag ==1 )
                        result.push_back(carryFlag);
                }
                else if(p1 == NULL && p2 != NULL)
                {
                    while(p2 != NULL)
                    {
                        curNum = (p2->val+carryFlag) %10;
                        if(p2->val + carryFlag >= 10)
                            carryFlag = 1;
                        else
                            carryFlag = 0;

                    result.push_back(curNum);
                    p2 = p2->next;
                    }
                    if(carryFlag == 1 )
                        result.push_back(carryFlag);
                }

            list<int>::iterator it = result.begin();
            for(;it != result.end(); it++)
                cout<<*it;
            return createListNode2(result);
        }
};

int main ()
{
    int arr1[] = {1};
    int arr2[] = {9,9};
    Solution s1;
    ListNode *l1 = createListNode(arr1,1);
    ListNode *l2 = createListNode(arr2,2);
    s1.addTwoNumbers(l1,l2);
    return 0;
    return 0;
}