24. Swap Nodes in Pairs [twenty-two switching node in the linked list]

description

Given a list pairwise exchange with adjacent nodes, and return the list to the exchange.
You can not just simply change the internal node values, but need to be the actual node exchange.

Examples of

ideas

• Recursive: to determine whether the node is empty, or only one node, otherwise, save second, not saved can not find it, if the press operation, finally, return to second
  
• Iteration
  
  answer
• python

*递归*
    def swapPairs(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return head
        second=head.next
        head.next=self.swapPairs(second.next)
        second.next=head
        
        return second
*迭代*
class Solution:
    def swapPairs(self, head: ListNode) -> ListNode:
        #因为要反转，所以给链表构造一个头结点
        newHead = ListNode(0)
        newHead.next = head
        cur=newHead 
        
        while cur.next and cur.next.next:
            # 因为改变链表的顺序，所以，先保存第一个不变的结点
            first = cur.next
            second = cur.next.next
            
            first.next=second.next
            second.next=first
            cur.next=second
            
        
            cur=cur.next.next
        #如果还有一个结点，不用反转了
        return newHead.next

• c++

*递归*
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        //如果为空，或者下一个结点为空，直接返回
        if (!head || !head->next)
            return head;
        
        ListNode* second = head->next;
        
        head->next = swapPairs(second->next);
        second->next = head;
        return second;
        
    }
};
*迭代*
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode* newHead = new ListNode(0);
        newHead->next = head;
        ListNode* cur=newHead;
        
        while(cur->next && cur->next->next)
        {
            ListNode *first=cur->next, *second=cur->next->next;
            first->next = second->next;
            second->next = first;
            cur->next = second;
            cur=cur->next->next;
            
        }

        return newHead->next;
    }
};