鏈表操作--- Remove Nth Node From End of List，注意事項整理

  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.

從尾端刪除第n個節點。

（1）獲取總的鏈表長度，L根據n帶入(L−n+1)求出從第一個節點開始數要刪除的節點。

public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode dummy = new ListNode(0);//1、在first節點前增加一個節點，防止僅有一個節點時有誤
    dummy.next = head;
    int length  = 0;
    ListNode first = head;
    while (first != null) {
        length++;//獲取所有節點的長度，包含了加入的判斷節點
        first = first.next;
    }
    length -= n;
    first = dummy;
    while (length > 0) {
        length--;
        first = first.next;//循環到要刪除節點的上一個節點
    }
    first.next = first.next.next;
    return dummy.next;//2、返回first節點
}

兩個while循環分別執行L，L-N次，時間複雜度O(L)；

（2）分別設定兩個指針，指向頭節點，第一個指針後移n-1，然後再通過第一個指針是否爲空，判斷第二個

指針節點位置，正好爲刪除節點位置。其實跟上邊通過length循環查找時一樣的，都做了length - n次循環

public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode first = dummy;
    ListNode second = dummy;
    // Advances first pointer so that the gap between first and second is n nodes apart
    for (int i = 1; i <= n + 1; i++) {
        first = first.next;
    }
    // Move first to the end, maintaining the gap
    while (first != null) {
        first = first.next;
        second = second.next;
    }
    second.next = second.next.next;
    return dummy.next;
}