019_删除链表的倒数第 N 个结点

中等

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode left = head;
        ListNode right = head.next;
        ListNode s = head;
        int count = 0;
        while(s != null){
            count++;
            s = s.next;
        }
        if(n == count){
            return head.next;
        }
        while(n < count-1){
            right = right.next;
            left = left.next;
            count--;
        }
        if(n == 1){
            left.next = null;
            return head;
        }else{
            left.next = right.next;
            return head;
        }
    }
}

先遍历一边链表，得出链表的总个数count，如果n=count，返回head.next，否则定义一个left和right指针，指针不断后移，count--，直到count==n，判断right后面是否有值，如果没有，left.next = null，否则left.next = right.next

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        dummy = ListNode(0, head)
        first = head
        second = dummy
        for i in range(n):
            first = first.next

        while first:
            first = first.next
            second = second.next
        
        second.next = second.next.next
        return dummy.next

快慢指针