19. Remove Nth Node From End of List

https://leetcode.com/problemset/all/?search=19

涉及链表删除操作的时候，稳妥起见都用 dummy
node，省去很多麻烦。因为不一定什么时候 head 就
被删了。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if (head == null) {
            return head;
        }
        int len = 0;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        while (head != null) {
            len++;
            head = head.next;
        }
        head = dummy;
        int index = len - n;
        while (index > 0) {
            head = head.next;
            index--;
        }
        head.next = head.next.next;
        return dummy.next;
    }
}

　快慢指针

public ListNode removeNthFromEnd(ListNode head, int n) {
    
    ListNode start = new ListNode(0);
    ListNode slow = start, fast = start;
    slow.next = head;
    
    //Move fast in front so that the gap between slow and fast becomes n
    for(int i=1; i<=n+1; i++)   {
        fast = fast.next;
    }
    //Move fast to the end, maintaining the gap
    while(fast != null) {
        slow = slow.next;
        fast = fast.next;
    }
    //Skip the desired node
    slow.next = slow.next.next;
    return start.next;
}