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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?
/**
* 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 node = new ListNode(0);
node.next = head;
ListNode fast = head;
ListNode slow = head;
while(n--!=-1){
fast = fast.next;
}
while(fast != null){
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return node.next;
}
}
使用了双指针,在做之前一定要想好特殊用例