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.
删除链表倒数第n个结点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n) {
ListNode*prev=NULL;
ListNode*p1=head;
ListNode*p2=head;
int k=0;
while(k<n && p2){k++;p2=p2->next;}
if(k<n)return head; //链表元素少于n个，则不做删除操作
//定位待待删除元素
while(p2){
prev=p1;
p1=p1->next;
p2=p2->next;
}
//删除元素
if(prev)prev->next=p1->next;
else head=p1->next;
return head;
}
};LeetCode: Remove Nth Node From End of List [019],布布扣,bubuko.com
LeetCode: Remove Nth Node From End of List [019]
原文地址：http://blog.csdn.net/harryhuang1990/article/details/25972443
