分类解决问题的思想

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

此题解决分为几个步骤，（整个问题的特殊解如：head==nulptr，head->next==nullptr）第一个一般性结点的解决，第二个特殊结点的解决（仔细分好），第三遍历所有的问题集

分类的思想应用=== 

然后在尾部情况应用break；跳出尾循环会减少对while的结束要求复杂度

class Solution {
public:
ListNode* swapPairs(ListNode* head) {
if(head == nullptr)return head;
if(head->next==nullptr)return head;
auto res = head->next;
while(head->next!=nullptr&&head!=nullptr)
{
auto headA = head;
if(head->next->next==nullptr){swapThree(headA);break;}
else if(head->next->next->next==nullptr){swapOne(headA);break;}
else{
head = head->next->next;
swapTwo(headA);
}
}
return res;
}
void swapOne(ListNode* H)
{
auto mark = H->next->next;
H->next->next = H;
H->next = mark;
mark->next = nullptr;
}
void swapTwo(ListNode* H)
{
auto mark = H->next->next;
H->next->next = H;
H->next = mark->next;
}
void swapThree(ListNode* H)
{
H->next->next = H;
H->next = nullptr;
}
};