LeetCode19- Remove Nth Node From End of List-Medium

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删除链表中倒数第n个结点

题目:LeetCode19

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:

Could you do this in one pass?

 

思路:

题目要求one pass,即只遍历一遍链表并完成删除操作。

如果删除倒数第n个结点,需要定位到该结点的前一个结点。

两种情况,1)删除的结点是头结点;2)删除的结点非头结点

首先:定义一个快指针,从链表头部走n步。如果此时已经指向null,说明要删除的结点是该链表的头结点,则返回head.next。即case 1。

case 2: 如果快指针此时没有指向null, 定义一个慢指针指向头结点,快慢指针同时向后移动,当快指针指向最后一个结点时,慢指针指向被删除结点的前一个。

操作慢指针,删除倒数第n个结点(slow.next = slow.next.next)。(注意:这里慢指针和快指针至少相差1,所以当快指针指向最后一个结点,slow.next.next 不会nullpointer异常。)

 

代码:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode fast = head;
        
        while (n > 0) {
            fast = fast.next;
            n--;
        }
        
        if (fast == null) {
            return head.next;
        }
        
        ListNode slow = head;
        while (fast.next != null) {
            slow = slow.next;
            fast = fast.next;
        }
        
        slow.next = slow.next.next;
        return head;
    }
}

 

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