【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;

     }

 };

巴特西

【LeetCode】024. Swap Nodes in Pairs

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