Add Two Numbers(转)

LeetCode之Add Two Numbers

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

【题意】

给你两个链表，表示两个非负整数。数字在链表中按反序存储，例如342在链表中为2->4->3。链表每一个节点包含一个数字（0-9）。

计算这两个数字和并以链表形式返回。

【分析】

无

【代码1】

 

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/********************************* *   日期：2014-01-27 *   作者：SJF0115 *   题号: Add Two Numbers *   来源：http://oj.leetcode.com/problems/add-two-numbers/ *   结果：AC *   来源：LeetCode *   总结： **********************************/ #include <iostream> #include <stdio.h> #include using namespace std;   struct ListNode {     int val;     ListNode *next;     ListNode(int x) : val(x), next(NULL) {} };   class Solution { public:     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {         ListNode *head = (ListNode *)malloc(sizeof(ListNode));         ListNode *pre = head;         ListNode *node = NULL;         //进位         int c = 0,sum;         //加法         while(l1 != NULL && l2 != NULL){             sum = l1->val + l2->val + c;             c = sum / 10;             node = (ListNode *)malloc(sizeof(ListNode));             node->val = sum % 10;             node->next = NULL;             //尾插法             pre->next = node;             pre = node;             l1 = l1->next;             l2 = l2->next;         }         //例如：2->4->3->1   5->6->4         while(l1 != NULL){             sum = l1->val + c;             c = sum / 10;             node = (ListNode *)malloc(sizeof(ListNode));             node->val = sum % 10;             node->next = NULL;             //尾插法             pre->next = node;             pre = node;             l1 = l1->next;         }         //例如：2->4->3   5->6->4->1         while(l2 != NULL){             sum = l2->val + c;             c = sum / 10;             node = (ListNode *)malloc(sizeof(ListNode));             node->val = sum % 10;             node->next = NULL;             //尾插法             pre->next = node;             pre = node;             l2 = l2->next;         }         //最后一位还有进位         if(c > 0){             node = (ListNode *)malloc(sizeof(ListNode));             node->val = c;             node->next = NULL;             //尾插法             pre->next = node;             pre = node;         }         return head->next;     } }; int main() {     Solution solution;     int A[] = {2,4,7,9};     int B[] = {5,6,4};     ListNode *head = NULL;     ListNode *head1 = (ListNode*)malloc(sizeof(ListNode));     ListNode *head2 = (ListNode*)malloc(sizeof(ListNode));     head1->next = NULL;     head2->next = NULL;     ListNode *node;     ListNode *pre = head1;     for(int i = 0;i < 4;i++){         node = (ListNode*)malloc(sizeof(ListNode));         node->val = A[i];         node->next = NULL;         pre->next = node;         pre = node;     }     pre = head2;     for(int i = 0;i < 3;i++){         node = (ListNode*)malloc(sizeof(ListNode));         node->val = B[i];         node->next = NULL;         pre->next = node;         pre = node;     }     head = solution.addTwoNumbers(head1->next,head2->next);     while(head != NULL){         printf("%d ",head->val);         head = head->next;     }     return 0; }   </algorithm></stdio.h></iostream>

 

【代码2】

 

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class Solution { public:     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {         ListNode *head = (ListNode *)malloc(sizeof(ListNode));         ListNode *pre = head;         ListNode *node = NULL;         //进位         int c = 0,sum,val1,val2;         //加法         while(l1 != NULL || l2 != NULL || c != 0){             val1 = (l1 == NULL ? 0 : l1->val);             val2 = (l2 == NULL ? 0 : l2->val);             sum = val1 + val2 + c;             c = sum / 10;             node = (ListNode *)malloc(sizeof(ListNode));             node->val = sum % 10;             node->next = NULL;             //尾插法             pre->next = node;             pre = node;             l1 = (l1 == NULL ? NULL : l1->next);             l2 = (l2 == NULL ? NULL : l2->next);         }         return head->next;     } };