Leetcode 24 ：Swap Nodes in Pairs

题目大意：给定一个链表，交换其中相邻的2个元素

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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/**
* 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) return NULL;

if(head != NULL && head->next == NULL) return head;

ListNode* headF = new ListNode(0);
headF->next = head;

ListNode* headO = headF;
while(headO->next != NULL && headO->next->next != NULL)
{
//
ListNode* tmpN1 = headO->next->next->next;

ListNode* tmpN = headO->next;

headO->next = headO->next->next;

headO->next->next = tmpN;

headO->next->next->next = tmpN1;

headO = headO->next->next;
}

ListNode*Re = headF->next;

delete headF;

return Re;
}
};

class Solution {
public:
ListNode* swapPairs(ListNode* head) {
bool first = true;
ListNode *pFirst = head, *pSecond = NULL, *pTail = NULL, *ret = head;
while (pFirst && pFirst -> next) {  // 判断是否存在两个后继元素
pSecond = pFirst -> next;       // 获取pSecond
if (first) ret = pSecond;       // 处理链表头
if (!first) pTail -> next = pSecond;    // 处理链表中段
pFirst -> next = pSecond -> next;
pSecond -> next = pFirst;       // 交换元素
pTail = pFirst;                 // 更新链表末尾

pFirst = pFirst -> next;        // 进行到下一个元素
first = false;
}
return ret;
}
};