24. Swap Nodes in Pairs

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.

题目：交换相邻的两个节点，如第一个和第二个交换，第三个和第四个交换。以此类推

-------更新------------2017-11-13 22:33:01

法一：

带个头节点比较好

class Solution {
public:
     ListNode* swapPairs(ListNode* head) {
        ListNode* dummy=new ListNode(0);
        dummy->next=head;
        ListNode* prev=dummy;
        while(head &&head->next)
        {
            ListNode* nn=head->next->next;
            prev->next=head->next;
            head->next->next=head;
            head->next=nn;
            prev=head;
            head=nn;
           
        }
        return dummy->next;
        
    }
};

 法二：官网上有人提交的一个方法。引用技巧，不如法一好理解..

ListNode* swapPairs(ListNode* head) {
        ListNode **pp = &head, *a, *b;
        while ((a = *pp) && (b = a->next)) {
            a->next = b->next;
            b->next = a;
            *pp = b;   
            pp = &(a->next);
        }
        return head;
    }

递归法： https://leetcode.com/problems/swap-nodes-in-pairs/discuss/

public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode newhd = head.next;
        head.next = swapPairs(newhd.next);
        newhd.next = head;
        return newhd;
}