【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;
    }
};