[leetcode][list] Swap Nodes in Pairs

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.

/** * 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(NULL == head || NULL == head->next) return head;        ListNode *headNew = head->next;//指向新链表的第一个节点        ListNode *pFirst = head;//指向本节点对的第一个节点        ListNode *pSecond = head->next;//指向本节点对的第二个节点        ListNode *pNext = head->next->next;//指向本节点对的下一个节点        ListNode *pLast = head;//指向新链表的最后一个节点        pSecond->next = pFirst;        while(pNext && pNext->next){            pFirst = pNext;            pSecond = pFirst->next;            pNext = pSecond->next;            pSecond->next = pFirst;            pLast->next = pSecond;            pLast = pFirst;        }        if(pNext) pLast->next = pNext;        else pLast->next = NULL;        return headNew;    }};

以下是无视题目中要求不能修改节点val的版本

/** * 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(NULL == head || NULL == head->next) return head;        ListNode *pOdd(head), *pEven(head->next);        while(pOdd && pOdd->next && pEven->next){//!!!先保证pOdd不为NULL撒            int iTmp = pOdd->val;//交换两个节点的val            pOdd->val = pEven->val;            pEven->val = iTmp;            pOdd = pOdd->next->next;//更新pOdd            pEven = pEven->next->next;//更新pEven        }        if(NULL == pOdd ->next) return head;//节点个数为奇数        //节点个数为偶数        int iTmp = pOdd->val;        pOdd->val = pEven->val;        pEven->val = iTmp;        return head;//节点没变，只是改变了节点的val，所以head没变    }};

递归实现：

/** * 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 (NULL == head || NULL == head->next) return head;ListNode *p = head->next;ListNode *q = p->next;p->next = head;head->next = swapPairs(q);return p;}};