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：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* swapPairs(ListNode* head) { 12 if(!head) return NULL; 13 if(!head->next) return head; 14 ListNode * helper1 = new ListNode(INT_MIN); 15 ListNode * helper2 = new ListNode(INT_MIN); 16 helper1->next = head; 17 helper2->next = head->next;// 先记录下首节点的位置,供函数返回的时候使用 18 ListNode * prev = head; 19 ListNode * curr = head->next; 20 while(prev){ 21 if(curr){ 22 ListNode * tmpNode = curr->next; 23 curr->next = prev; 24 helper1->next = curr; 25 prev->next = tmpNode; 26 helper1 = prev; 27 if(prev->next && prev->next->next){ 28 prev = prev->next; 29 curr = prev->next; 30 }else 31 break; 32 }else 33 break; 34 35 } 36 return helper2->next; 37 } 38 };
