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
1 #include "stdafx.h" 2 #include <malloc.h> 3 #include <iostream> 4 using namespace std; 5 typedef struct ListNode { 6 int val; 7 ListNode *next; 8 ListNode(int x) : val(x), next(NULL) {} 9 10 }*Lnode; 11 Lnode create() //尾插法建立链表 12 { 13 Lnode head,q,r; 14 int temp; 15 head = (struct ListNode *)malloc(sizeof(Lnode)); 16 head->next = NULL; 17 r = head; 18 cin >> temp; 19 while (temp!=-1)//输入-1，创建链表结束 20 { 21 q = (struct ListNode *)malloc(sizeof(Lnode)); 22 q->val = temp; 23 r->next = q; 24 r = q; 25 cin >> temp; 26 } 27 r->next = NULL; 28 return head; 29 } 30 void print(Lnode h) //打印链表 31 { 32 Lnode p=h->next; 33 while (p) 34 { 35 cout << p->val<<endl; 36 p = p->next; 37 } 38 } 39 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) 40 { 41 int flag = 0; 42 ListNode * p,*r; 43 ListNode * h = (struct ListNode *)malloc(sizeof(ListNode *)); 44 h->next = NULL; 45 r = h; 46 int num; 47 if (l1 == NULL) return l1; 48 if (l2 == NULL) return l2; 49 while (l1&&l2) 50 { 51 num = l1->val + l2->val + flag; 52 if (num<10) 53 { 54 p = (struct ListNode *)malloc(sizeof(ListNode *)); 55 p->val = num; 56 r->next = p; 57 r = p; 58 flag = 0; 59 } 60 else //如果两位数相加大于10，则向前进一位 61 { 62 p = (struct ListNode *)malloc(sizeof(ListNode *)); 63 p->val = num-10; 64 r->next = p; 65 r = p; 66 flag = 1; 67 } 68 l1 = l1->next; 69 l2 = l2->next; 70 } 71 while (l1) 72 { 73 num = l1->val + flag; 74 if (num<10) 75 { 76 p = (struct ListNode *)malloc(sizeof(ListNode *)); 77 p->val = num; 78 r->next = p; 79 r = p; 80 flag = 0; 81 } 82 else 83 { 84 p = (struct ListNode *)malloc(sizeof(ListNode *)); 85 p->val = num - 10; 86 r->next = p; 87 r = p; 88 flag = 1; 89 } 90 l1 = l1->next; 91 92 } 93 94 while (l2) 95 { 96 num = l2->val + flag; 97 if (num<10) 98 { 99 p = (struct ListNode *)malloc(sizeof(ListNode *)); 100 p->val = num; 101 r->next = p; 102 r = p; 103 flag = 0; 104 } 105 else 106 { 107 p = (struct ListNode *)malloc(sizeof(ListNode *)); 108 p->val = num - 10; 109 r->next = p; 110 r = p; 111 flag = 1; 112 } 113 l2 = l2->next; 114 115 } 116 if (flag) //最后再判断一次判断是否有进位 117 { 118 p = (struct ListNode *)malloc(sizeof(ListNode *)); 119 p->val = 1; 120 r->next = p; 121 r = p; 122 } 123 r->next = NULL; 124 return h->next; 125 } 126 int _tmain(int argc, _TCHAR* argv[]) //测试函数 127 { 128 Lnode t1,t2,t; 129 t1 = create(); 130 t2 = create(); 131 t = addTwoNumbers(t1->next,t2->next); 132 print(t); 133 return 0; 134 }
Java版本：
public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { if(l1==null) return l2; if(l2==null) return l1; ListNode head=new ListNode(0); head.next=null; int tag=0;//进位标志 if(l1.val+l2.val<10) { head.val=l1.val+l2.val; } else { head.val=l1.val+l2.val-10; tag=1; } ListNode p=head; ListNode node1=l1.next; ListNode node2=l2.next; while(node1!=null||node2!=null) { int sum; int result; ListNode node=new ListNode(0); node.next=null; p.next=node; p=node; if(node1!=null&&node2!=null) { sum=node1.val+node2.val; result=sum+tag; if(result<10) { node.val=result; tag=0; } else { node.val=result-10; tag=1; } node1=node1.next; node2=node2.next; } else if(node1==null&&node2!=null) { sum=node2.val; result=sum+tag; if(result<10) { node.val=result; tag=0; } else { node.val=result-10; tag=1; } node2=node2.next; } else if(node1!=null&&node2==null) { sum=node1.val; result=sum+tag; if(result<10) { node.val=result; tag=0; } else { node.val=result-10; tag=1; } node1=node1.next; continue; } } if(node1==null&&node2==null&&tag==1) { ListNode node=new ListNode(1); node.next=null; p.next=node; p=node; } return head; } }