Add Two Numbers

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#C++ #LeetCode #链表

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本文介绍了一种使用链表表示非负整数并进行加法运算的方法。通过遍历两个链表，逐位相加，并处理进位问题，最终返回表示两数之和的新链表。文章提供了一个C++实现示例。

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题目描述：

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.

思路:就是两个数的加法，唯一要注意的是最高位之后还有进位要将其加上。

#include<cstdio>
struct ListNode 
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) 
{
    //结果链表的头节点 
    ListNode* result = NULL;
    //结果链表的游标 
    ListNode* pointer = NULL;
    int num1 = 0;
    int num2 = 0;
    //进位数 
    int addition = 0;
	while(l1!=NULL||l2!=NULL)
	{
		num1 = (l1==NULL)?(0):(l1->val);
		num2 = (l2==NULL)?(0):(l2->val);
		//该位节点的数值 
		int val = (num1+num2+addition)%10;
		//更新下一位的进位数 
		addition = (num1+num2+addition)/10;
		if(result==NULL)
		{
			result = new ListNode(val);
			pointer = result; 
		}		    
		else
		{
			ListNode* node = new ListNode(val);
			pointer->next = node;
			pointer = node;
		} 
		if(l1!=NULL) 
		    l1 = l1->next;
		if(l2!=NULL)
		    l2 = l2->next;  
	}	
	//如果最高位计算完之后还有进位，要将其加上 
	if(addition!=0)
	{
		ListNode* node = new ListNode(addition);
		pointer->next = node;
	}		    
	return result; 
}

int main(){
	ListNode* node1 = new ListNode(5);
	ListNode* node2 = new ListNode(4);
	ListNode* node3 = new ListNode(3);
	node1->next = node2;
	node2->next = node3;
	
	ListNode* node4 = new ListNode(5);
	ListNode* node5 = new ListNode(6);
	ListNode* node6 = new ListNode(4);
	ListNode* node7 = new ListNode(4);
	node4->next = node5;
	node5->next = node6;
	node6->next = node7;
	ListNode* result = addTwoNumbers(node1,node4);
	while(result!=NULL)
	{
		printf("%d",result->val);
		if(result->next!=NULL)
		    printf("->"); 
		result = result->next;
	}
	return 0;
}