蜗牛慢慢爬 LeetCode 19. Remove Nth Node From End of List [Difficulty: Medium]

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.

翻译

注意是要移除从链表结尾数的第n个

Hints

Related Topics: LinkedList, Two Pointers
通过两个指针 后一个指针在前一个指针移动 n-1 个之后再移动 就能保证该指针是要移除的那个了

代码

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        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;
    }
}

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
        """
        t = ListNode(head.val)
        t.next = head
        a = b = c = t
        count = 0
        while a.next!=None:
            a = a.next
            count += 1
            if count>=n:
                c = b
                b = b.next
        c.next = b.next
        head = t.next
        return head