Remove Nth Node From End of List
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题目:
Given a linked list, remove the nth node from the end of list and return its head.
For example,
Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.
题意删除链表中倒数第n个节点,不改变链表其他节点结构,返回链表头指针。
由于链表是单向的,不能倒着搜索,因此设置两个指针,他们相隔n个节点的距离,同布移动。
当前面的指针到达链表结尾的时候,后面一个指针离结尾的距离正好是n,再删除要求的节点就ok了。
C语言版:
struct ListNode* removeNthFromEnd(struct ListNode* head, int n)
{
struct ListNode* post=head;
struct ListNode* pre=head;
struct ListNode* temp=NULL;
if(head==NULL)
return NULL;
int i;
for(i=0;i<n-1;i++)
pre=pre->next;
while(pre->next)
{
temp=post;
pre=pre->next;
post=post->next;
}
if(temp==NULL)
{
head=head->next;
}
else{
temp->next=post->next;
}
return head;
}
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