445. Add Two Numbers II**

445. Add Two Numbers II**

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

题目描述

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

C++ 实现 1

先用 Reverse 的方法. 之后再考虑不修改输入节点的方法.

class Solution {
private:
    ListNode* reverse(ListNode *head) {
        ListNode *pre = nullptr;
        auto ptr = head;
        while (ptr) {
            ListNode *tmp = ptr->next;
            ptr->next = pre;
            pre = ptr;
            ptr = tmp;
        }
        return pre;
    }
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        l1 = reverse(l1);
        l2 = reverse(l2);

        int carry_over = 0;
        ListNode *dummy = new ListNode(0);
        auto ptr = dummy;
        while (l1 || l2 || carry_over) {
            int part1 = l1 ? l1->val : 0;
            int part2 = l2 ? l2->val : 0;

            int sum = carry_over + part1 + part2;
            ptr->next = new ListNode(sum % 10);
            ptr = ptr->next;
            carry_over = sum / 10;

            l1 = l1 ? l1->next : nullptr;
            l2 = l2 ? l2->next : nullptr;
        }
        return reverse(dummy->next);
    }
};

C++ 实现 2

想不修改输入节点, 需要使用栈.

class Solution {
private:
	// 传递指针是可以修改原节点的
    void insert(ListNode *head, int val) {
        ListNode *node = new ListNode(val);
        node->next = head->next;
        head->next = node;
    }
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        stack<int> s1, s2;
        while (l1) {
            s1.push(l1->val);
            l1 = l1->next;
        }
        while (l2) {
            s2.push(l2->val);
            l2 = l2->next;
        }

        int carry_over = 0;
        ListNode *dummy = new ListNode(0);
        while (!s1.empty() || !s2.empty() || carry_over) {
            int part1 = 0, part2 = 0;
            if (!s1.empty()) {
                part1 = s1.top();
                s1.pop();
            }
            if (!s2.empty()) {
                part2 = s2.top();
                s2.pop();
            }

            int sum = part1 + part2 + carry_over;
            insert(dummy, sum % 10);
            carry_over = sum / 10;
        }
        return dummy->next;
    }
};