LeetCode OJ：Swap Nodes in Pairs（成对交换节点）

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.

如题实例所示，需要成对的交换节点，这里我用到了两个帮助节点，一个记下起点的前面位置，一个作为每一对prev的前面一个节点使用，思路比较简单，就是交换之后再向后面移动两个节点就可以了，代码如下所示：

/**
 * 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) return NULL;
        if(!head->next) return head;
          ListNode * helper1 = new ListNode(INT_MIN);
          ListNode * helper2 = new ListNode(INT_MIN);
          helper1->next = head;
          helper2->next = head->next;// 先记录下首节点的位置,供函数返回的时候使用
          ListNode * prev = head;
          ListNode * curr = head->next;
          while(prev){
              if(curr){
                  ListNode * tmpNode = curr->next;
                  curr->next = prev;
                  helper1->next = curr;
                  prev->next = tmpNode;
                  helper1 = prev;
                  if(prev->next && prev->next->next){
                      prev = prev->next;
                      curr = prev->next;
                  }else
                      break;
              }else
                  break;
              
          }      
          return helper2->next;
    }
};