题解 | #链表相加(二)#

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

class Solution {
public:
    /**
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    ListNode* reverseList(ListNode* head){//链表翻转
        auto L=new ListNode(0);
        ListNode* p=head,*q=nullptr;
        while(p){
            q=p->next;
            p->next=L->next;
            L->next=p;
            p=q;
        }
        return L->next;
    }

    ListNode* addInList(ListNode* head1, ListNode* head2) {
        // write code here
        ListNode *L1=reverseList(head1);
        ListNode *L2=reverseList(head2);

        ListNode *L3=L1;
        ListNode *pre=NULL;//断点的前驱
        int res=0;
        int jinwei=0;
        int yushu=0;

        while(L1&&L2){
            res=L1->val+L2->val+jinwei;
            jinwei=res/10;
            yushu=res%10;
            L1->val=yushu;
            pre=L1;
            L1=L1->next;
            L2=L2->next;
        }

        while(L1){//L2完成
            res=L1->val+jinwei;
            jinwei=res/10;
            yushu=res%10;
            L1->val=yushu;
            pre=L1;
            L1=L1->next;
        }

        if(L2){//L1完成
            pre->next=L2;
            while(L2){
                res=L2->val+jinwei;
                jinwei=res/10;
                yushu=res%10;
                L2->val=yushu;
                pre=L2;
                L2=L2->next;
            }
        }
        
        if(jinwei){//最后进位
            ListNode* s=new ListNode(1);
            pre->next=s;
        }

        return reverseList(L3);
    }
};