Swap Nodes in Pairs[medium]

Description

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.

Analysis

題目是很容易理解的。就是將相鄰的結點交換，並且題目中明確要求不能直接交換結點的值，同時，從題目中我們能夠看出他是每兩個結點進行一次交換。我採用的方法是每三個結點交換後兩個，首先建立一個結點h，它的next指標指向list的head，如果h的下一個和下下個均不為空，我們即可利用h來交換後兩個結點，之後更新h的值，是他等於上一次處理的最後一個結點。

Solution

/**
 * 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* h = new ListNode(1);
        h->next = head;
        ListNode* c = h;
        ListNode* p = NULL;
        ListNode* t = NULL;
        while (c->next != NULL
 
 &&c->next->next != NULL) {
            p = c->next;
            t = c->next->next;
            p->next = t->next;
            c->next = t;
            t->next = p;
            c = p;
        }
        return h->next;
    }
};