description:

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

You may not modify the values in the list's nodes, only nodes itself may be changed.
Note:

Example:

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

my answer:

感恩

画图，没有难点，就是两个两个来回挪，要设置pre指向要挪的两个节点的前面那个节点, t指向要挪的两个节点的后面那个节点。

大佬的answer：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode* dummy = new ListNode(-1), *pre = dummy;
        dummy->next = head;
        while(pre->next && pre->next->next){
            ListNode* t = pre->next->next;
            pre->next->next = t->next;
            t->next = pre->next;
            pre->next = t;
            pre = t->next;
        }return dummy->next;
    }
};

relative point get√：

hint :