Swap Nodes in Pairs

来源：互联网发布：originpro数据拟合编辑：程序博客网时间：2024/05/16 06:54

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.

思路：题目要求交换链表的相邻的两个节点，但是不许修改节点的数据，只是移动节点的位置。可以给这个链接加一个头节点然后进行处理，代码如下：

/** * 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 *res=new ListNode(0),*pre=res,*cur;        pre->next=head;        cur=head;        while(cur&&cur->next)        {            pre->next=cur->next;            cur->next=cur->next->next;            pre->next->next=cur;            pre=cur;            cur=cur->next;        }        return res->next;            }};

方法二：在discuss上看到有使用递归的方式，可以参考下

class Solution {public:    ListNode* swapPairs(ListNode* head) {        if(head == NULL||head->next==NULL)            return head;        ListNode* next = head->next;        head->next = swapPairs(next->next);        next->next = head;        return next;    }};

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