1.题目 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.
2.解法 思路：快慢指针 时间复杂度：O(N)
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { // 2 pointers ListNode start = new ListNode(0); ListNode slow = start, fast = start; slow.next = head; for(int i = 0; i <= n; i++){ fast = fast.next; } while(fast != null){ slow = slow.next; fast = fast.next; } slow.next = slow.next.next; return start.next; } }