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.
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/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */public
class Solution { /**We use two pointer to to do this: * 1.move the first point to n-th node of the linked list * 2.add a new head to the linked list and put the second point on this new pointer * 3. move both pointer forward, until the next node of the first pointer is null. * 4. remove the next of second pointer. * 5. get rid of the fake head. * * @param head -- ListNode, head node of a linked list * @param n --Integer, remove the n-th node from the end of the list * @return ListNode, the head of the modified linked list * @author Averill Zheng * @version 2014-06-05 * @since JDK 1.7 */ public
ListNode removeNthFromEnd(ListNode head, int
n) { if(n == 0) return
head; ListNode tail = head; for(int
i = 1; i < n; ++i) tail = tail.next; ListNode newHead = new
ListNode(0); newHead.next = head; ListNode beforeRemovedNode = newHead; while(tail.next != null){ tail = tail.next; beforeRemovedNode = beforeRemovedNode.next; } beforeRemovedNode.next = beforeRemovedNode.next.next; newHead = newHead.next; return
newHead; }} |
leetcode--Remove Nth Node From End of List,布布扣,bubuko.com
leetcode--Remove Nth Node From End of List
原文地址：http://www.cnblogs.com/averillzheng/p/3773945.html
