# 引言
题目链接：https://leetcode.com/problems/add-two-numbers/description/
# 题目大意
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)
# 题解
# 一句话题解
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)
# AC代码
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
}
有什么想法，欢迎留下评论。