19. Remove Nth Node From End of List
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.
题目大意：
找到链表中倒数第N个元素，删除这个元素。
代码如下：
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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
:
int
lengthOfList(ListNode* head)
{
int
i = 0 ;
while
(head != NULL)
{
i++;
head = head->next;
}
return
i;
}
ListNode* removeNthFromEnd(ListNode* head,
int
n) {
if
(head == NULL)
return
NULL;
ListNode* p = head;
int
pre = lengthOfList(head) - n ;
if
(pre == 0)
return
head->next;
cout << pre<<
" "
<<lengthOfList(head)<<endl;
while
(--pre)
p = p->next;
p->next = p->next->next;
return
head;
}
};
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本文转自313119992 51CTO博客，原文链接:http://blog.51cto.com/qiaopeng688/1837285
