題目19. Remove Nth Node From End of List

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.

1

思路:   
使用兩個快慢指針,一個slow,一個fast
首先fast前進n個節(jié)點
然后slow從頭(head)和fast一起每次前進一個節(jié)點,當fast鏈表尾部的時候,
slow指向的就是倒數第n個節(jié)點 

public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode fast = head;
        for (int i = 0; i < n; i++) {
            fast = fast.next;
        }
        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;
    }
}