19. Remove Nth Node From End of List

题目描述

Given a linked list, remove the nth node from the end of

For example,

Note:
Given n will always

解题方法

1. Consider that head是否會被remove , 因為有可能，所以create dummy head.
2. 因為是從end nth remove, index 從1開始, 所以利用two pointers 從dummy node 走nth node到要被移除的element 前面， 然後two pointers 同步走到底，其中tail pointer 就會走到要被移除的element前面
3. Then only manipulate the tail pointer to remove the specified element

    public ListNode removeNthFromEnd(ListNode head, int n){
        if(head == null || head.next == null || n <= 0){return null;}

        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode fast = dummy;
        ListNode slow = dummy;
        for(int i = 0; i < n; i++){
            if(fast == null){
                return null;
            }
            fast = fast.next;
        }
        while(fast.next != null){
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;
        return dummy.next;
    }