24. Swap Nodes in Pairs 解题记录

题目描述：

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.

解题思路：

把head前再加上一个节点pre代表要换位置的两个节点的前一个节点，然后循环遍历整个链表就行了。

代码：

/**
 * 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* pre = new ListNode(0);
        pre->next = head;
        ListNode* h = pre;
        while(pre->next&&pre->next->next){
            ListNode* t = pre->next;
            pre->next = t->next;
            t->next = t->next->next;
            pre->next->next=t;
            pre = pre->next->next;
        }
        return h->next;
    }
};