[LeetCode] 19. 删除链表的倒数第N个节点

题目链接:https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/

题目描述:

给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。

示例:

给定一个链表: 1->2->3->4->5, 和 n = 2.

当删除了倒数第二个节点后，链表变为 1->2->3->5.

思路:

使用快慢指针,

快指针先移n个节点

接下来,快慢指针一起移动,两指针之间一直保持n个节点,当快指针到链表底了,操作慢指针,删除要删除的元素!

时间复杂度:(O(n))

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代码:

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

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        if not head:return 
        dummy = ListNode(0)
        dummy.next = head
        fast = dummy
        while n:
            fast = fast.next
            n -= 1
        slow = dummy
        while fast and fast.next:
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return dummy.next

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 dummy = new ListNode(0);
        dummy.next = head;
        ListNode slow = dummy;
        ListNode fast = dummy;
        for (int i = 0 ; i < n; i ++ ){
            fast = fast.next;
        }
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next;
        }
        slow.next = slow.next.next;
        return dummy.next;
        
    }
}