Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:

Could you do this in one pass?

 
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode pSlow = head;
ListNode pFast = head;
for(int i = 0; i < n; i++){
pFast = pFast.next;
} if(pFast == null){ //remove first node
return head.next;
} while(pFast.next!=null){
pFast = pFast.next;
pSlow = pSlow.next;
}
pSlow.next = pSlow.next.next;
return head; }
}

解题思路:使用快慢指针,实现O(n)遍历。

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