【LeetCode】024. 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.

题解：

　　没什么好写的，链表题一般要注意头结点的问题，还有一定要画图！

Solution 1

 /**
  * 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 dummy(-);
         dummy.next = head;
         ListNode* cur = &dummy;
 
         while (cur->next && cur->next->next) {
             ListNode* tmp = cur->next->next;
             cur->next->next = tmp->next;
             tmp->next = cur->next;
             cur->next = tmp;
             cur = tmp->next;
         }
 
         return dummy.next;
     }
 };

　　递归：

Solution 2

 /**
  * 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 || !head->next)
             return head;
         ListNode* newHead = head->next;
         head->next = swapPairs(head->next->next);
         newHead->next = head;
         return newHead;
     }
 };