LeetCode 2. Add Two Numbers C++-优快云博客

本文链接：https://blog.youkuaiyun.com/mooe1011/article/details/83718847

本文介绍了一种使用链表结构来实现两个非负整数相加的算法。通过将数字存储在链表中，每个节点包含一个数字，且数字以逆序排列。算法遍历两个链表，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.

#include<stdio.h>
using namespace std;

// Definition for singly-linked list.
struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}	
};

class Solution {
public:
	ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
		int digit = 0, carry = 0;
		ListNode* res = new ListNode(0), *p = res;
		while (l1 || l2 || carry) {
			digit = carry;
			if (l1) {
				digit += l1->val;
				l1 = l1->next;
			}
			if (l2) {
				digit += l2->val;
				l2 = l2->next;
			}
			carry = digit / 10;
			if (digit > 9) {				
				digit %= 10;
			}
			p->next = new ListNode(digit);
			p = p->next;
		}
		return res->next;
	}	
};

ListNode* creatList(int a[], int size) {
	ListNode* res = new ListNode(0), *p = res;
	for (int i = 0; i < size; i++) {
		p->next = new ListNode(a[i]);
		p = p->next;
	}
	p->next = NULL;
	return res->next;
}

void printList(ListNode* L) {
	ListNode *p = L;
	while (p->next!=NULL) {
		printf("%d -> ", p->val);
		p = p->next;
	}
	printf("%d", p->val);
}

int main() {
	int L1[] = { 2,4,3 };
	int L2[] = { 5,6,4 };
	ListNode* l1 = creatList(L1, sizeof(L1) / sizeof(int));
	ListNode* l2 = creatList(L2, sizeof(L2) / sizeof(int));
	ListNode* l3 = Solution().addTwoNumbers(l1, l2);
	printList(l3);
	return 0;
}