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：
/**
* 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;
}
};2016-08-12 14:02:00
本文出自 “做最好的自己” 博客，请务必保留此出处http://qiaopeng688.blog.51cto.com/3572484/1837285
leetCode 19. Remove Nth Node From End of List 链表
原文：http://qiaopeng688.blog.51cto.com/3572484/1837285