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
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。
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/********************************** 日期：2014-01-27* 作者：SJF0115* 题号: 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> |
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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; }};
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原文：http://www.cnblogs.com/rat-bin/p/4361943.html