Swap Nodes in Pairs--LeetCode

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：将链表拆分，然后再合并

List *swapPairs(List *head) {    if(head == NULL || head->next == NULL)     return head;    List* ji=head;    List* ou = head->next;    List* temp1,*temp2;    List* l1= head;    List* l2 = head->next;    temp1 = l2->next;    while(temp1 != NULL)    {      temp2 = temp1->next;      l1->next = temp1;      l2->next = temp2;      l1 = temp1;      l2 = temp2;      if(l2 !=NULL)      temp1 = l2->next;      else      temp1 = NULL;    }    if(l1 != NULL)    l1->next =NULL;    if(l2 != NULL)    l2->next = NULL;    temp1 = ou;    while(temp1 != NULL&&temp1->next != NULL)    {        temp2 = ji->next;       ji->next = temp1->next;       temp1->next = ji;       temp1 = ji->next;       ji = temp2;             }    temp1->next = ji;    return ou;} 

上面的思路是将一个链表拆分然后再合并，这里仍然使用一个新的节点来作为指引节点来操作。

List* SwapNode(List* list){List* head=new List;head->next = list;List* cur = head;List* pre,*next;while(cur->next != NULL && cur->next->next != NULL){pre = cur->next;next = cur->next->next;cur->next = next;next = next->next;cur->next->next = pre;pre->next = next;cur = pre;}return head->next;}

ps:其实问题的思路非常的简单，但是解决问题的方法决定了处理问题的简易程度，比如我们始终使用一个指针的next域来指向准备反转的一对指针的第二个指针，然后再反转这对指针即可，这里使用一个伪指针来处理这个问题非常的合适。