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Given a linked list, remove the nth node from the end of list and return its head.
Given linked list: 1->2->3->4->5->null, and n = 2.
/** * Definition for ListNode. * public class ListNode { * int val; * ListNode next; * ListNode(int val) { * this.val = val; * this.next = null; * } * } */ public class Solution { /** * @param head: The first node of linked list. * @param n: An integer. * @return: The head of linked list. */ ListNode removeNthFromEnd(ListNode head, int n) { // write your code here ListNode fakehead = new ListNode(0); fakehead.next = head; ListNode fast = fakehead; for(int i = 0; i <= n; i++) fast = fast.next; ListNode slow = fakehead; while(fast != null){ slow = slow.next; fast = fast.next; } slow.next = slow.next.next; return fakehead.next; } }
lintcode-easy-Remove Nth Node from End of List
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原文地址:http://www.cnblogs.com/goblinengineer/p/5246616.html