leetcode -- 19. Remove Nth Node From End of List

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题目描述

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?

AC代码

/*** 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 slow = head,  fast = head;for(int i = 0;i < n;i++) {fast = fast.next;if(fast == null && i == n - 1) return head.next; //删除最后一个节点的情况}while(fast.next != null){fast = fast.next;slow = slow.next;}slow.next = slow.next.next;return head;}
}

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