C++两个单链表相加生成一个单链表进阶解法 空间复杂度O(1)

#include <iostream>


struct Node {
int value;
Node *next;
Node(int data) : value(data), next(nullptr) {};
};


Node *reverse(Node * head);
Node *addTwoList(Node * head1, Node *head2);
void print(Node *head);


int main()
{
Node n1(9);
Node n2(3);
Node n3(8);
n1.next = &n2;
n2.next = &n3;


Node b1(6);
Node b2(3);
b1.next = &b2;


//print(reverse(&n1));// 测试链表翻转函数以及print函数
print(addTwoList(&n1, &b1));


system("pause");
return 0;
}


Node * reverse(Node * head) {
if (head == nullptr) {
return head;
}


Node * pre = nullptr;
Node * next = nullptr;
while (head != nullptr) {
next = head->next;
head->next = pre;
pre = head;
head = next;
}
return pre;
}


void print(Node * head)
{
while (head != nullptr) {
std::cout << head->value << std::endl;
head = head->next;
}
}


Node *addTwoList(Node * head1, Node *head2)
{
head1 = reverse(head1);
head2 = reverse(head2);


int add = 0, n1 = 0, n2 = 0;
Node * cur1 = head1, *cur2 = head2, *n = nullptr;


while (cur1 != nullptr || cur2 != nullptr) {
n1 = cur1 == nullptr ? 0 : cur1->value;
n2 = cur2 == nullptr ? 0 : cur2->value;


if (cur1!= nullptr)
cur1 = cur1->next;
if (cur2 != nullptr)
cur2 = cur2->next;


Node * node = new Node((n1 + n2 + add) % 10);
if (n1 + n2 + add >= 10)
add = 1;
else
add = 0;


node->next = n;
n = node;
}
if (add == 1) {
Node * node = new Node(1);
node->next = n;
n = node;
}


reverse(head1);
reverse(head2);
return n;
}