【leetcode】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.

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一A的水题，哦啦啦啦啦~

就注意一点：不能光顾着该当前的两个链表节点的指针，还要用last_pair存放上一组最后一个节点的指针，以修改这个节点的next指针指向当前组新的第一个指针。如果没有上一组时候要特殊处理head指针。有点饶，举个例子，题目给的例子里面，当交换1，2的时候注意head指针要指向2而不是1；交换完1，2得到如下链表2->1->3->4，当交换3，4的时候注意要修改1的next指针指向4。

代码如下：

#include <iostream>
using namespace std;
struct ListNode {
    int val;
    ListNode *next;
     ListNode(int x) : val(x), next(NULL) {}
 };

class Solution {
public:
    ListNode *swapPairs(ListNode *head) {
        ListNode* current = head;
        ListNode* last_pair = head;
        while(current != NULL){
            ListNode* temp = current->next;
            if(temp != NULL){
                if(last_pair == head)
                    head = temp;
                else
                    last_pair->next = temp;
                last_pair = current;    
                current->next = temp->next;
                temp->next = current;
            }

            current = current->next;
        }
        return head;
    }
};

int main(){
    Solution s;
    ListNode* head = new ListNode(1);
    ListNode* node1= new ListNode(2);
    head->next = node1;

    ListNode* node2= new ListNode(3);
    node1->next = node2;

    ListNode* node3= new ListNode(4);
    node2->next = node3;

    ListNode* r = s.swapPairs(head);
    while(r != NULL){
        cout <<r->val<<endl;
        r = r->next;
    }
}

 JAVA版本：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
        if(head == null)
            return head;
        
        ListNode newHead = null;
        ListNode one = head;
        ListNode two = head.next;
        ListNode before = null;
        
        while(one != null && two != null){
            ListNode three = two.next;
            two.next = one;
            one.next = three;
            
            if(newHead == null)
            {
                newHead = two;
                before = one;
            }else {
                before.next = two;
                before = one;
            }  
            one = three;
            if(one != null)
                two = one.next;
        }
        if(one != null && before != null)
            before.next = one;
        return newHead==null?head:newHead;
    }
}