LeetCode刷題之第一題——Add Two Numbers
阿新 • • 發佈:2019-02-01
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
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ListNode* p1=l1,*p2=l2; ListNode* result=NULL,*temp=NULL,*presult=NULL;//result為結果連結串列頭,temp為每位上相加後的值,presult為將temp連結到result上 int up=0; while(p1!=NULL&&p2!=NULL){ temp= new ListNode(p1->val+p2->val); temp->val=up+temp->val;//先加上進位得到本位之和 up=temp->val/10; //得到本位之和的進位值 temp->val=temp->val%10;//得到本位值 if(result==NULL){//空連結串列第一次賦值 result=temp; presult=result; } else{ presult->next=temp; presult=temp; } p1=p1->next; p2=p2->next; } while(p2!=NULL){ temp= new ListNode(p2->val); temp->val=up+temp->val; up=temp->val/10; temp->val=temp->val%10; if(result==NULL){ result=temp; presult=result; } else{ presult->next=temp; presult=temp; } p2=p2->next; } while(p1!=NULL){ temp= new ListNode(p1->val); temp->val=up+temp->val; up=temp->val/10; temp->val=temp->val%10; if(result==NULL){ result=temp; presult=result; } else{ presult->next=temp; presult=temp; } p1=p1->next; } if(up>0)//[5] [5]結果為[0][1]而不是[0] { temp= new ListNode(up); presult->next=temp; } return result; } };