19. Remove Nth Node From End of List

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.

链接:  http://leetcode.com/problems/remove-nth-node-from-end-of-list/

题解：

one pass remove Nth node from the end。要建立一个dummy node，用来记录所要移除节点之前的节点，再利用先后指针就可以找到这个节点了。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if(head == null)
            return head;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode first = dummy;      //because n is always valid, so we don't need to get the length of the linked list and do mod to n
        ListNode second = dummy;
        
        while(first.next != null){
            first = first.next;
            n--;
            if(n < 0)
                second = second.next;
        }
        
        second.next = second.next.next;
        return dummy.next;
    }
}

 

二刷:

要计算好是哪一个节点。

/**
 * 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 dummy = new ListNode(-1);
        dummy.next = head;
        ListNode fast = dummy, slow = dummy;
        while (fast.next != null) {
            fast = fast.next;
            if (n <= 0) {
                slow = slow.next;    
            }
            n--;
        }
        slow.next = slow.next.next;
        return dummy.next;
    }
}

Python:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        dummy = ListNode(-1)
        dummy.next = head
        slow = dummy
        fast = dummy
        while fast.next != None:
            fast = fast.next
            if n <= 0:
                slow = slow.next
            n -= 1
        slow.next = slow.next.next
        return dummy.next

三刷: 

用个例子就很容易能算出来是哪个节点了。  List = {1, 2, 3}, n = 2

Java:

/**
 * 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 dummy = new ListNode(-1);
        dummy.next = head;
        ListNode slow = dummy, fast = dummy;
        while (fast.next != null) {
            fast = fast.next;
            if (n <= 0) {
                slow = slow.next;
            }
            n--;
        }
        slow.next = slow.next.next;
        return dummy.next;
    }
}

Reference:

https://leetcode.com/discuss/37149/3-short-python-solutions