Swap Nodes in Pairs

问题描述：

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.

问题分析：

将链表中每一对结点对换，采用递归求解，直接模拟就可以

问题求解：

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

/** * 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 == NULL || head->next == NULL){            return head;        }                ListNode* newHead = new ListNode(0);        newHead->next = head;        ListNode* preNode = newHead;        ListNode* curNode = head;                while(curNode != NULL && curNode->next != NULL)        {            // swap curNode and curNode->next            preNode->next = curNode->next;            curNode->next = preNode->next->next;            preNode->next->next = curNode;  //此处我写成 curNode = preNode->next->next, 编译不通过，对链表的操作还是有些不清楚                        // go over two nodes            preNode = curNode;            curNode = curNode->next;        }        head = newHead->next;        delete newHead;        return head;    }};