2.Add Two Numbers(2)

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后来又重新整理了，与（1）中相比，（1）中只有两个循环，但是（2）中有三个三循环，因此时间也从22ms变了28ms.

但是（2）相比（1）节省了空间，特别是在这个两个数的位数很大，且两个数的位数很近的时候。

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {    struct ListNode* sumNode=NULL; //Use struct to define the variable    struct ListNode* head=NULL;        struct ListNode* one=l1;    struct ListNode* two=l2;        struct ListNode* list=NULL;    struct ListNode* another=NULL;        if(l1==NULL && l2==NULL)    {        return two;    }    else if(l1==NULL)    {        return l2;    }    else if(l2==NULL)    {        return l1;    }    else //two lists are not NULL    {        //Find which list is longer        //set list to the shorter list        while(one!=NULL && two!=NULL)        {            one=one->next;            two=two->next;        }        if(one!=NULL)        {            list=l2; //shouldn't set to two            another=l1;        }        else        {            list=l1;            another=l2;        }                struct ListNode* pre;        int num=0;                while(list!=NULL)        {            if(head==NULL)            {                head=another;            }                        int sum=list->val+another->val+num;            num=0;                        if(sum<10)            {                another->val=sum;            }            else            {                int remain=sum%10;                num=sum/10;                another->val=remain;            }                        pre=another;            list=list->next;            another=another->next;        }                while(num!=0 && another!=NULL)        {            int sum=another->val+num;            num=0;                        if(sum<10)            {                another->val=sum;            }            else            {                int remain=sum%10;                num=sum/10;                another->val=remain;            }                        pre=another;            another=another->next;                    }                if(num!=0 && another==NULL)        {            sumNode=(struct ListNode*)malloc(sizeof(struct ListNode*));            sumNode->val=num;            sumNode->next=NULL;                        pre->next=sumNode;        }            }        return head;}

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