【leetcode 445】Add Two Numbers II

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:

What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:


Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 8 -> 0 -> 7

要求时间复杂度 O(n),空间复杂度 O(1).

思路：

1. 空间复杂度O(1)就必须在原有链表的基础上进行运算。因此需要选择出长度较大的链表来储存最后的结果。

2.时间复杂度O(n)就只能从后往前遍历。单链表从后往前遍历，首先想到反转，因此单向链表反转的代码一定要非常熟练。

3.单向链表反转，需要定义一个头结点 dummy，然后将除链表第一个结点之外的结点，依次插入到头结点之后（头插法），然后返回 dummy.next.【非常重要】。

#include <iostream>
#include <string>
using namespace std;

struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
ListNode(int x, ListNode* l) : val(x), next(l) {}
};

ListNode* digitToList(string N){
ListNode* first = new ListNode(0);
ListNode* current = first;
for (auto c : N){
current->next = new ListNode(c - '0');
current = current->next;
}
return first->next;
}

ListNode* reverse(ListNode* head){
if (head == nullptr || head->next == nullptr) return head;
ListNode dummy(-1);
dummy.next = head;
ListNode *prev = head, *cur = prev->next;
while (cur != nullptr){
prev->next = cur->next;
cur->next = dummy.next;
dummy.next = cur;
cur = prev->next;
}
return dummy.next;
}

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2){
ListNode* rl1 = reverse(l1);
ListNode* rl2 = reverse(l2);

int len1=0, len2=0;
for (ListNode* temp=rl1; temp != nullptr; temp = temp->next) len1++;
for (ListNode* temp = rl2; temp != nullptr; temp = temp->next) len2++;
ListNode* storage = len1 >= len2 ? rl1 : rl2;
ListNode* other = storage == rl1 ? rl2 : rl1;
ListNode* p = nullptr;
for (p = storage; p != nullptr; p = p->next){
if (other != nullptr){
p->val += other->val;
other = other->next;
}
}
int carry = 0;
for (p = storage; p != nullptr; p = p->next){
p->val += carry;
carry = p->val / 10;
p->val %= 10;
}
ListNode* rstorage = reverse(storage);
if (carry) {
ListNode *newHead = new ListNode(carry);
newHead->next = rstorage;
return newHead;
}
return rstorage;
}

int main(){
string A, B;
getline(cin, A);
getline(cin, B);
ListNode *l1 = digitToList(A), *l2 = digitToList(B);
ListNode* res = addTwoNumbers(l1, l2);
while (res){
cout << res->val;
res = res->next;
}
cout << endl;
}