leetcode 2 Add Two Numbers

最新推荐文章于 2022-02-22 21:07:37 发布

原创最新推荐文章于 2022-02-22 21:07:37 发布 · 326 阅读

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本文介绍了一种使用链表表示非负整数并实现两数相加的方法。通过遍历两个链表，逐位进行加法运算，并处理进位情况，最终返回结果链表。文章详细展示了算法的具体实现过程。

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

/**

 * Definition for singly-linked list.

 * struct ListNode {

 *     int val;

 *     ListNode *next;

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

 * };

 */
#include <vector>
#include <cstdlib>
#include <cstdio>
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 = NULL;
		ListNode* tail = NULL;
		ListNode* p = l1;
		ListNode* q = l2;
		int carry = 0;
		while(p != NULL && q != NULL)
		{
			ListNode* new_v = (ListNode*)malloc(sizeof(ListNode));
			new_v->next = NULL;
			if(p->val + q->val + carry >= 10)
			{
				new_v->val = p->val + q->val + carry -10;
				carry = 1;
			}	
			else
			{
				new_v->val = p->val + q->val + carry;
                                carry = 0;
			}
			if(head == NULL)
			{
				head = new_v;
				tail = new_v;
			}
			else
			{
				tail->next = new_v;
				tail = new_v;
			}
			p = p->next;
			q = q->next;	
		}
		while(p != NULL)
		{
			ListNode* new_v = (ListNode*)malloc(sizeof(ListNode));
                        new_v->next = NULL;
                        if(p->val + carry >= 10)
                        {
                                new_v->val = p->val + carry -10;
                                carry = 1;
                        }
                        else
                        {
                                new_v->val = p->val + carry;
                                carry = 0;
                        }
                        if(head == NULL)
                        {
                                head = new_v;
                                tail = new_v;
                        }
                        else
                        {
                                tail->next = new_v;
				tail = new_v;
                        }
                        p = p->next;
		}
		while(q != NULL)
		{
			ListNode* new_v = (ListNode*)malloc(sizeof(ListNode));
                        new_v->next = NULL;
                        if(q->val + carry >= 10)
                        {
                                new_v->val = q->val + carry -10;
                                carry = 1;
                        }
                        else
                        {
                                new_v->val = q->val + carry;
                                carry = 0;
                        }
                        if(head == NULL)
                        {
                                head = new_v;
                                tail = new_v;
                        }
                        else
                        {
                                tail->next = new_v;
				tail = new_v;
                        }
                        q = q->next;
		}
		if(carry == 1)
		{
			ListNode* new_v = (ListNode*)malloc(sizeof(ListNode));
                        new_v->next = NULL;
			new_v->val = 1;
			if(head == NULL)
                        {
                                head = new_v;
                                tail = new_v;
                        }
                        else
                        {
                                tail->next = new_v;
				tail = new_v;
                        }	
		
		}
		return head;

  	}

};

int main()
{
	ListNode* l1 = NULL, *l2 = NULL;
	ListNode* t1 = NULL, *t2 = NULL;
	int a, b;
	scanf("%d", &a);
	for(int i = 1; i <= a; i++)
	{
		int x;
		scanf("%d", &x);
		ListNode* new_p = (ListNode*)malloc(sizeof(ListNode));
		new_p->next = NULL;
		new_p->val = x; 

		if(l1 == NULL)
		{
			l1 = new_p;
			t1 = new_p;
		}		
		else
		{
			t1->next = new_p;
		}
	}
	scanf("%d", &a);
        for(int i = 1; i <= a; i++)
        {
                int x;
                scanf("%d", &x);
                ListNode* new_p = (ListNode*)malloc(sizeof(ListNode));
                new_p->next = NULL;
                new_p->val = x;
        
                if(l2 == NULL)
                {
                        l2 = new_p;
                        t2 = new_p;
                }
                else
                {
                        t2->next = new_p;
                }
        }	

	Solution s;
	ListNode* r = s.addTwoNumbers(l1, l2);
	
	while(r != NULL)	
	{
		printf("%d ", r->val);
		r = r->next;
	}
	return 0;

}

读题一定要仔细。

还有题目没有说两个输入链表长度相等，就要考虑不相等的情况。

还有最后一位需要进位。