标签:des style blog http io color ar os sp
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
注意:最后一位如果有进位,需要新创建一个结点来存放。
C++代码实现:
#include<iostream> #include<new> using namespace std; //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 *p=l1; ListNode *pre=l1; ListNode *q=l2; int carry=0; while(p&&q) { if(p->val+q->val+carry>9) { p->val=(p->val+q->val+carry)%10; carry=1; } else { p->val+=q->val+carry; carry=0; } pre=p; p=p->next; q=q->next; } if(q) pre->next=q; while(q) { if(q->val+carry>9) { q->val=(q->val+carry)%10; carry=1; } else { q->val+=carry; carry=0; break; } pre=q; q=q->next; } while(p) { if(p->val+carry>9) { p->val=(p->val+carry)%10; carry=1; } else { p->val+=carry; carry=0; break; } pre=p; p=p->next; } if(carry) { p=new ListNode(1); pre->next=p; } return l1; } void createList(ListNode *&head,int *arr) { ListNode *p=NULL; int i=0; for(i=0; i<5; i++) { if(head==NULL) { head=new ListNode(arr[i]); if(head==NULL) return; } else { p=new ListNode(arr[i]); p->next=head; head=p; } } } }; int main() { Solution s; ListNode *L1=NULL; ListNode *L2=NULL; ListNode *L=NULL; int arr1[10]= {1,9,7,5,3}; int arr2[10]= {0,8,6,4,2}; s.createList(L1,arr1); s.createList(L2,arr2); L=s.addTwoNumbers(L1,L2); while(L) { cout<<L->val<<" "; L=L->next; } }
运行结果:
标签:des style blog http io color ar os sp
原文地址:http://www.cnblogs.com/wuchanming/p/4100195.html