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19. Remove Nth Node From End of List

时间:2015-05-13 08:46:52      阅读:101      评论:0      收藏:0      [点我收藏+]

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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.

双指针,第一个指针比第二个指针快n个节点。注意边界情况。

/**
* 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) {
    ListNode tail = head;
    while (n > 0) {
      // No need to delete
      if (tail == null) {
        return head;
      }
      tail = tail.next;
      n--;
    }
    // The deleted one is head
    if (tail == null) {
      return head.next;
    }
    ListNode tmp = head;
    while (tail.next != null) {
      tmp = tmp.next;
      tail = tail.next;
    }
    tmp.next = tmp.next.next;
    return head;
  }
}

19. Remove Nth Node From End of List

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原文地址:http://www.cnblogs.com/shini/p/4499285.html

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