Given a linked list, swap every two adjacent nodes and return its head.
For example,Given1->2->3->4, you should return the list as2->1->4->3.
Your algorithm should use only constant space. You maynotmodify the values in the list, only nodes itself can be changed.
题意：两两交换结点
思路：首先很容易用两个指针t1，t2做到前后结点的交换，，然后再用一个结点来记录前一个，然后就能处理了
/** * 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) {if (head == NULL) return NULL;if (head->next == NULL) return head;ListNode *t1, *t2, *pre;t1 = head;pre = head;while (t1 != NULL && t1->next != NULL) {t2 = t1->next;t1->next = t2->next;t2->next = t1;if (pre != head)pre->next = t2;else head = t2;pre = t1;t1 = t1->next;}return head;}};
相信梦想是价值的源泉，相信眼光决定未来的一切，
