"Leetcode" Swap Nodes in Pairs (3 solutions)

Swap Nodes in Pairs

Given a linked list, swap every, 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. Modify the values in the list, only nodes itself can be changed.

Solution One: Recursion

/** Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution { Public: ListNode*swappairs (ListNode *head) {        if(head = = NULL | | head->next = =NULL)returnHead; ListNode* Newhead = head->Next; Head->next = Swappairs (newhead->next); Newhead->next =Head; returnNewhead; }};

Solution Two: Reverse Nodes in K-group order k=2

/** Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution { Public: ListNode*swappairs (ListNode *head) {        returnReversekgroup (Head,2); } ListNode*reversekgroup (ListNode *head,intk) {ListNode* Newhead =NewListNode (-1); ListNode* Tail =Newhead; ListNode* begin =Head; ListNode* end =begin;  while(true)        {            intCount =K;  while(Count && End! =NULL) {End= end->Next; Count--; }            if(Count = =0)            {//reverse from [begin, end]Stack<listnode*>s;  while(Begin! =end)                    {S.push (begin); Begin= begin->Next; }                 while(!S.empty ()) {ListNode* top =S.top ();                    S.pop (); Tail->next =top; Tail= tail->Next; }            }            Else            {//Leave outTail->next =begin;  Break; }        }        returnNewhead->Next; }};

Solution Three:

After each pair of nodes 22 is exchanged, a connection is returned to the previous node.

/** Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution { Public: ListNode*swappairs (ListNode *head) {ListNode* Newhead =NewListNode (-1); Newhead->next =Head; ListNode* cur =Newhead;  while(Cur->next! = NULL && Cur->next->next! =NULL) {cur->next = Reverse (Cur->next, cur->next->next); Cur= cur->next->Next; }        returnNewhead->Next; } ListNode* Reverse (listnode* N1, listnode*n2) {//swap adjacent nodes, the latter is returnedN1->next = n2->Next; N2->next =N1; returnN2; }};

"Leetcode" Swap Nodes in Pairs (3 solutions)