Swap Nodes in Pairs

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.

題目要求鏈表不能修改，只能修改Node，這樣反而使得題目變的簡單了；有興趣的可以嘗試下，只能修改鏈表，不能修改node，也就是說要兩兩反轉鏈表，而且也是在常數空間，以及O(n)的時間完成。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    //修改node，不修改List
    void Swap(int &a, int &b)
    {
        int tmp = a;
        a = b;
        b = tmp;
    }
    
    ListNode* swapPairs(ListNode* head) {
        if(head == NULL)
            return NULL;
        ListNode *tmpNode = head;
        while(tmpNode != NULL && tmpNode->next != NULL)
        {
            Swap(tmpNode->val, tmpNode->next->val);
            tmpNode = tmpNode->next->next;
        }
        
        return head;
    }
};

直接反轉指針

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    //兩兩反轉節點
    ListNode* swapPairs(ListNode* head) {
        if(head == NULL)
            return NULL;
        ListNode *curNode = head, *bNode = head, *nextNode = NULL;
        while(curNode != NULL && curNode->next != NULL)
        {
            nextNode = curNode->next;
            if(curNode == head)
            {
                head = nextNode;
            }
            else
            {
                bNode->next = nextNode;
            }
            
            curNode->next = nextNode->next;
            nextNode->next = curNode;
            bNode = curNode;
            curNode = curNode->next;
        }
        
        return head;
    }
};