【LeetCode】两数相加

1.主要是链表先创建下一个对象，再转移，而不是先转移到空指针再赋值

t = new ListNode((v1 + v2 + count1)%10);
count1 = (v1+v2+count1)/10;
t = t->next;
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t->next = new ListNode((v1 + v2 + count1)%10);
count1 = (v1+v2+count1)/10;
t = t->next;

#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;

 struct ListNode {
      int val;
      ListNode *next;
      ListNode() : val(0), next(nullptr) {}
      ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
  };


    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* head = nullptr; ListNode* t = nullptr;
        int count1 = 0;
        while (l1 || l2) 
        {
            int v1 = l1 ? l1->val : 0;
            int v2 = l2 ? l2->val : 0;
            if(!head) 
            {
                t = head = new ListNode((v1 + v2)%10);
                count1 = (v1+v2)/10;
            }else
            {
                t->next = new ListNode((v1 + v2 + count1)%10);
                count1 = (v1+v2+count1)/10;
                t = t->next;
            }
            if(l1)
            l1 = l1->next;
            if(l2)
            l2 = l2->next;
        }
        if(count1 != 0)
        {
            t->next = new ListNode(count1);
        }
        return head;
    }

int main()
{
	ListNode* head =new ListNode(2);
	ListNode* v1 = head;
	head->next =new ListNode(4);
	head = head->next;
	head->next =new ListNode(3);
	ListNode* head2 =new ListNode(5);
	ListNode* v2 = head2;
	head2->next =new ListNode(6);
	head2 = head2->next;
	head2->next =new ListNode(4);
	ListNode* v3 =  addTwoNumbers(v1,v2);
	int a = 0;
	a = 1;
}

2.三目运算符

条件是否为真？ 是 ： 否

3.初始化

ListNode* a = nullptr;