【LeetCode】2. Add Two Numbers（C++）_add two numbers c++-优快云博客

本文链接：https://blog.youkuaiyun.com/Ethan95/article/details/84072854

本文解析了LeetCode上两数相加的链表问题，通过逆序链表表示数字并求和，提供了两种解决方案：一是直接修改第一个链表，二是创建新链表返回结果。代码详细展示了如何处理进位和遍历链表。

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地址：https://leetcode.com/problems/add-two-numbers/

题目：

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.

理解：

两个链表，逆序表示两个非负的数字，要求求两个数字的和。
我的实现把l1作为返回结果的链表。每次算一位保存在l1，然后记录进位，直到某个链表遍历完了。如果遍历完的是l1，把l2剩余的补到l1的后面。然后把多的借位加到另一个链表即可。

实现：

class Solution {
public:
	ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
		int carry = 0;
		ListNode *p1, *p2;
		ListNode *head = l1;
		while (l1&&l2) {
			int tmp = l1->val + l2->val + carry;
			l1->val = tmp % 10;
			carry = tmp / 10;
			p1 = l1;
			p2 = l2;
			l1 = l1->next;
			l2 = l2->next;
		}
		if (l2) {
			p1->next = l2;
			l1 = l2;
		}
		while (carry||l1) {
			if (l1) {
				int tmp = l1->val + carry;
				l1->val = tmp % 10;
				carry = tmp / 10;
				p1 = l1;
				l1 = l1->next;
			}
			else {
				p1->next = new ListNode(1);
				break;
			}
		}
		return head;
	}
};

看了下别人的代码，重新建一个链表整体的思路似乎更清楚一点。重新实现了一下。

class Solution {
public:
	ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
		int carry = 0;
		ListNode *p1, *p2;
		ListNode *head=new ListNode(0),*phead=head;
		while (l1||l2) {
			int tmp = carry;
			if (l1) {
				tmp += l1->val;
				p1 = l1;
				l1 = l1->next;
			}
			if (l2) {
				tmp += l2->val;
				p2 = l2;
				l2 = l2->next;
			}
			phead->next = new ListNode(tmp % 10);
			carry = tmp / 10;
			phead = phead->next;
		}
		if (carry) {
			phead->next = new ListNode(1);
		}
		return head->next;
	}
};

不知道这样为什么比上面的快一些。去洗澡。