Add Two Numbers

最新推荐文章于 2024-05-07 14:18:04 发布

sf_jiang

最新推荐文章于 2024-05-07 14:18:04 发布

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分类专栏： LeetCode刷题文章标签： add two numbers 链表 LeetCode

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本文介绍了一种使用链表实现加法运算的方法，特别适用于逆序存储的非负整数。通过遍历两个链表，逐位相加并处理进位，最终返回结果链表。

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You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

问题描述：从左到右加法计算，满10进位.

思路：很简单，三个变量表示进位值step(0,1)，当前的目的链表中的存入值value，以及sum. 注意的是最后要判断step是否为1，如果为1还要再创建一个节点。

附加代码：

#include<iostream>
#include<vector>
#include<string>
#include<map>
#include<unordered_map>
#include<stdio.h>
using namespace std;

struct ListNode {
   int val;
   ListNode *next;
   ListNode(int x) : val(x), next(NULL) {}
};

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
   int step = 0, sum = 0, value = 0;
   ListNode *p = new ListNode(0);
   ListNode *phead = p;
   //遍历
   while (l1 != NULL && l2 != NULL)
   {
       sum = l1->val + l2->val + step;
       step = sum / 10;
       value = sum % 10;
       ListNode *pnode = new ListNode(value);
       //更新目的链表
       p->next = pnode;
       p = pnode;
       //更新原链表
       l1 = l1->next;
       l2 = l2->next;
   }
   while (l1 != NULL)
   {
       sum = l1->val + step;
       step = sum / 10;
       value = sum % 10;
       ListNode *pnode = new ListNode(value);
       p->next = pnode;
       p = pnode;
       l1 = l1->next;
   }
   while (l2 != NULL)
   {
       sum = l2->val + step;
       step = sum / 10;
       value = sum % 10;
       ListNode *pnode = new ListNode(value);
       p->next = pnode;
       p = pnode;
       l2 = l2->next;
   }
   if (step != 0)
   {
       ListNode *pnode = new ListNode(step);
       p->next = pnode;
   }
   return phead->next;
}

void printList(ListNode *pl)
{
   while (pl != NULL)
   {
       cout << pl->val << " ";
       pl = pl->next;
   }
   cout << endl;
}

ListNode *creatList(vector<int> vec)
{
   ListNode *p = new ListNode(0);
   ListNode *phead = p;
   for (int i = 0; i < vec.size(); ++i)
   {
       ListNode *ptmp = new ListNode(vec[i]);
       p->next = ptmp;
       p = ptmp;
   }
   return phead->next;
}

int main()
{
   vector<int> vec = { 3,5,7 };
   ListNode *p = creatList(vec);
   printList(p);

   vector<int> vec1 = { 3,5,7 };
   ListNode *p1 = creatList(vec1);
   printList(p1);

ListNode *p2 = addTwoNumbers(p, p1);
printList(p2);

system("pause");
return 0;
}