第19题 Remove Nth Node From End of List

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本文介绍了一种高效算法，用于从链表中删除倒数第N个节点，并返回修改后的链表头。该算法仅遍历一次链表，通过两个指针策略实现目标节点的精确定位。

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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.

Solution：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode fakeHead = new ListNode(0);
        fakeHead.next =head;
        ListNode first =fakeHead, second =fakeHead;
        for(int i=0; i<n; i++){
            second=second.next;
        }
        while(second.next!=null){
            first=first.next; second=second.next;
        }
        first.next=first.next.next;
        return fakeHead.next;
    }
}