leetcode： 单链表之Add Two Numbers

leetcode： 单链表之Add Two Numbers

题目：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。

即：给定两个非负数，它们是按照逆序存储的，每个节点值保留一个数值，要求输出这两个数之和，返回结果链表。

C++实现：

#include <iostream>using namespace std;struct ListNode{int val;ListNode *next;ListNode (int x):val(x),next(NULL){ }};ListNode *createListNode(int arr[],int n){ListNode *r;ListNode *p;ListNode * L=(ListNode*)malloc(sizeof(ListNode));r=L;for(int i=0;i<n;i++){p=(ListNode*)malloc(sizeof(ListNode));p->val=arr[i];r->next=p;r=p;}r->next=NULL;    return L->next;}ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {     ListNode head(-1); // 头节点     int carry = 0;     ListNode *prev = &head;     for (ListNode *pa = l1, *pb = l2;          pa != NULL || pb != NULL;          pa = pa == NULL ? NULL : pa->next,          pb = pb == NULL ? NULL : pb->next,          prev = prev->next)  { const int ai = pa == NULL ? 0 : pa->val;         const int bi = pb == NULL ? 0 : pb->val;         const int value = (ai + bi + carry) % 10;         carry = (ai + bi + carry) / 10;         prev->next = new ListNode(value); // 尾插法     }     if (carry > 0)         prev->next = new ListNode(carry);     return head.next;}int main(){int a[]={2,4,3};int b[]={5,6,4};ListNode *a1 = createListNode(a,3);ListNode *b1 = createListNode(b,3);/*while(a1 != NULL){        cout<<a1->val;        a1 = a1->next;    }cout<<endl;*/ListNode *head = NULL;head=addTwoNumbers(a1,b1);while(head != NULL){        cout<<head->val;        head = head->next;    }cout<<endl;return 0;}

输出结果：