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.
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* swapPairs(struct ListNode* head)
{
struct ListNode *p=head,*q=head;
int LiskLen=0;
while(q!=NULL)
{
LiskLen++;
q=q->next;
}
if(LiskLen%2==0)
{
while(p!=NULL)
{
int temp;
temp=p->val;
p->val=p->next->val;
p->next->val=temp;
p=p->next->next;
}
}
else
{
while(p->next!=NULL)
{
int temp;
temp=p->val;
p->val=p->next->val;
p->next->val=temp;
p=p->next->next;
}
}
return head;
}
本文介绍了一种链表操作技巧,即交换每个相邻节点的值。通过迭代遍历链表,实现节点值的互换,同时保持链表的结构不变。此方法仅使用常量空间,不修改节点值,只改变节点间的连接。
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