#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.

解法：先画图搞清楚指针变换的关系，然后再改。这道题真他妈难，写不粗来，参考了别人的代码。巧妙地用了一个中间变量。

/**
 * 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*  current = head;
        ListNode* pnext = NULL;
        ListNode* pre = NULL;
        ListNode* prePre = NULL;
        while(current != NULL && current->next  != NULL) {
            pre = current;
            current = current->next;
            pnext = current->next;
            if(pre == head) head = current;    //这是首次进行循环头指针的改变
            if(prePre) prePre->next = current;  //这步是关键步骤，一开始没想到过
            
            current->next = pre;
            pre->next = pnext;
 
            prePre = pre;   //保存前指针
            current = pnext;
        }
        return head; 
    }
};