AddTwoSum

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {      //对应结构体初始化为ListNode(int data=0,struct ListNode next=NULL):val(val),next(next){}
        val = x;
        next = null;
    }
}

class Solution5 {
    ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode preHead = new ListNode(0);
        ListNode p = l1, q = l2, curr = preHead;
        int carry = 0;
        while (p != null || q != null) {
            int x = (p != null) ? p.val : 0;
            int y = (q != null) ? q.val : 0;
            int sum = x + y + carry;                //进位处理，并在下一个节点处进行处理
            carry = sum / 10;
            curr.next = new ListNode(sum % 10); //新构造一个节点
            curr = curr.next;
            if (p != null) p = p.next;
            if (q != null) q = q.next;
        }
        if (carry > 0) {                            //处理多一位
            curr.next = new ListNode(carry);
        }
        return preHead.next;
    }
}

此题基本就是用链表处理大数，所以用int+ListNode的方式来处理大数问题，只是应该找到其适用场景，譬如较多的写操作、内存占用等。

#include <bits/stdc++.h>
#include <map>
#include <vector>
using namespace std;

struct ListNode
{
    int val;
    struct ListNode *next;
    ListNode(int data = 0, struct ListNode *next = NULL) : val(val), next(next) {} //初始化链表
};

class Solution
{
  public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
    {
        ListNode preHead(0), *p = &preHead;
        int carry = 0;
        while (l1 || l2)
        {
            int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry;
            carry = sum / 10;
            p->next = new ListNode(sum % 10);
            p = p->next;
            l1 = l1 ? l1->next : l1;
            l2 = l2 ? l2->next : l2;
        }
        if (carry > 0)
        { //处理多一位
            p->next = new ListNode(carry);
        }
        return preHead.next;
    }