【Swap Nodes in Pairs】cpp

题目：

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.

代码：

/**

 * 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 dummy(-);

            dummy.next = head;

            ListNode *prev = &dummy;

            ListNode *curr = head;

            while ( curr && curr->next )

            {

                prev->next = curr->next;

                curr->next = curr->next->next;

                prev->next->next = curr;

                prev = curr;

                curr = curr->next;

            }

            return dummy.next;

    }

};

Tips:

链表基本操作，动手画图，直接出来。

===================================

第二次过这道题，一次AC了。

/**

 * 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 dummpy(-);

            dummpy.next = head;

            ListNode* pre = &dummpy;

            ListNode* curr = head;

            while ( curr && curr->next )

            {

                pre->next = curr->next;

                curr->next = curr->next->next;

                pre->next->next = curr;

                pre = curr;

                curr = curr->next;

            }

            return dummpy.next;

    }

};

巴特西

【Swap Nodes in Pairs】cpp

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