LeetCode_2: Add Two Numbers - LeetCode

# 引言

题目链接：https://leetcode.com/problems/add-two-numbers/description/

给出两个的倒序的链表，链表每个节点的值为非负个位数，链表反续串起来是一个数字，按照加法规则将两个链表相加并返回一个倒序链表，链表每个节点的数字对应计算结果的每一位。

例：

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

Explanation: 342 + 465 = 807.

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

(不含前导零，数字本身可以为零)

倒序链接正好由低位到高位表示了数字的每一位，直接取节点数据相加插入新链表节点即可 (注意进位问题)

复杂度 O(n)

c++ 版本

	class Solution
	{
	public:
	ListNode addTwoNumbers(ListNode l1, ListNode *l2)
	{
	ListNode *rList = NULL;
	ListNode *curNode;
	// 考虑进位
	int carry = 0;
	while (NULL != l1 \|\| NULL != l2)
	{
	int var1 = (NULL != l1) ? l1->val : 0;
	int var2 = (NULL != l2) ? l2->val : 0;
	if (NULL == rList)
	{
	rList = new ListNode((var1 + var2 + carry) % 10);
	curNode = rList;
	}
	else
	{
	curNode->next = new ListNode((var1 + var2 + carry) % 10);
	curNode = curNode->next;
	}
	l1 = (NULL == l1) ? l1 : l1->next;
	l2 = (NULL == l2) ? l2 : l2->next;
	carry = (var1 + var2 + carry) / 10;
	}
	if (carry > 0)
	{
	curNode->next = new ListNode(carry);
	}
	return rList;
	}
	};

go 版本 (感觉写的好挫，一点都不优雅 :[笑哭]:)

	func addTwoNumbers(l1 ListNode, l2 ListNode) *ListNode {
	var rList, curNode *ListNode
	var carry, var1, var2 int
	for nil != l1 \|\| nil != l2 {
	if nil != l1 {
	var1 = l1.Val
	l1 = l1.Next
	} else {
	var1 = 0
	}
	if nil != l2 {
	var2 = l2.Val
	l2 = l2.Next
	} else {
	var2 = 0
	}
	if nil == rList {
	rList = &ListNode{(var1 + var2 + carry) % 10, nil}
	curNode = rList
	} else {
	curNode.Next = &ListNode{(var1 + var2 + carry) % 10, nil}
	curNode = curNode.Next
	}
	carry = (var1 + var2 + carry) / 10
	}
	if carry > 0 {
	curNode.Next = &ListNode{carry, nil}
	}
	return rList
	}