【LeetCode】33.Linked List — Add Two Numbers两数相加

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

思路一：定义一个新链表。设置两个指针分别指向两个链表的开头。将相加的结果记录在新链表中。

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode*n1=l1;
        ListNode*n2=l2;
        ListNode*head=new ListNode(-1);//定义新链表的链表头
        ListNode*tag=head;//作为新链表的遍历指针
        int counter=0;    //用来记录进位情况
        while(n1!=NULL&&n2!=NULL) //同时遍历两个链表，当有一个到头时循环结束
        {
            tag->next=new ListNode(-1);
            tag=tag->next;
            tag->next=NULL;
            int i=n1->val+n2->val+counter;
            if(i>=10)
            {
                counter=1;
                tag->val=i-10;
                n1=n1->next;
                n2=n2->next;
            }
            else
            {
                counter=0;
                tag->val=i;
                n1=n1->next;
                n2=n2->next;
            }
            
        }
        if(n1==NULL&&n2!=NULL)  //考虑L2比L1长的情况
        {
            if(counter==0)      //考虑无进位的情况
            {
                ListNode*D=head;
                head=head->next;
                delete D;
                tag->next=n2;
                return head;
            }
            else{               // 考虑有进位的情况
                  while(n2!=NULL)
                  {
                      tag->next=new ListNode(-1);
                      tag=tag->next;
                      tag->next=NULL;
                      int j=n2->val+counter;
                      if(j>=10) //考虑有进位的情况
                      {
                         counter=1;
                         tag->val=j-10;
                         n2=n2->next;
                      }
                      else
                      {        //考虑无进位的情况     
                          counter=0;
                          tag->val=j;
                          ListNode*D=head;
                          tag->next=n2->next;
                          head=head->next;
                          delete D;
                          return head;
                       }
                   }
                ListNode *last=new ListNode(1); 
                tag->next=last;
                ListNode*D=head;
                head=head->next;
                delete D;
             }
        }
        else if(n1!=NULL&&n2==NULL)//考虑L1比L2长的情况
        {
            if(counter==0) //考虑无进位的情况
            {
                ListNode*D=head;
                head=head->next;
                delete D;
                tag->next=n1;
                return head;
            }
            else{// 考虑有进位的情况
                  while(n1!=NULL)
                  {
                      tag->next=new ListNode(-1);
                      tag=tag->next;
                      tag->next=NULL;
                      int j=n1->val+counter;
                      if(j>=10)//考虑有进位的情况
                      {
                         counter=1;
                         tag->val=j-10;
                         n1=n1->next;
                      }
                      else
                      {//考虑无进位的情况    
                          counter=0;
                          tag->val=j;
                          ListNode*D=head;
                          tag->next=n1->next;
                           head=head->next;
                          delete D;
                          return head;
                       }
                   }
                ListNode *last=new ListNode(1);
                tag->next=last;
                ListNode*D=head;
                head=head->next;
                delete D;
             }
        }
        else //考虑两个链表等长的情况
        {
            if(counter==0)  //考虑无进位的情况
            {
                ListNode*D=head;
                head=head->next;
                delete D;
                return head;
            }
            else  //考虑有进位的情况
            {
                ListNode *last=new ListNode(1);
                tag->next=last;
                ListNode*D=head;
                head=head->next;
                delete D;
                return head;
            }
        }  
        return head;
    }
};

思路二：第一种方法较为繁琐。思路二想到先遍历两个链表获取链表长度，再更具不同长度情况分别讨论。讨论不同情况时还要注意是否有进位。