LeetCodeOJ2 Add Two Sum

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;}