删除链表的倒数第N个结点

leetCode-19（删除链表的倒数第N个结点）

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

示例 1：

输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]

示例 2：

输入：head = [1], n = 1
输出：[]

示例 3：

输入：head = [1,2], n = 1
输出：[1]

解法一，自己瞎写

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode node = head;
        int count = 0;
        //计算链表中所有节点的数量
        while (node!=null){
            node = node.next;
            count++;
        }
        //特殊情况的判断
        if (count==0) return head;
        if (count==1) return null;
        node = head;
        //如果删除的节点是倒数第count个节点，则直接删除第一个即可
        if (count-n==0) {
            head = head.next;
            return head;
        }
        //定位到倒数第几个节点的前一个节点
        for (int i = 1;i<count-n;i++){
            node = node.next;
        }
        //删除节点操作
        node.next = node.next.next;
        node = head;
        return node;
    }
}

解法二：栈

代码为：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0, head);//定义辅助节点，防止删除头结点
        Stack<ListNode> stack = new Stack<>();
        ListNode node = dummy;
        while (node!=null){
            stack.push(node);
            node = node.next;
        }
        for (int i = 1;i<=n;i++){
            stack.pop();
        }
        ListNode pre = stack.peek();
        pre.next = pre.next.next;
        return dummy.next;
    }
}

不用辅助节点：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        Stack<ListNode> stack = new Stack<>();
        ListNode node = head;
        while (node!=null){
            stack.push(node);
            node = node.next;
        }
        for (int i = 1;i<=n;i++){
            stack.pop();
        }
        if (stack.size()==0) return head.next;//判断一下是否删除的是头结点
        node = stack.pop();
        node.next = node.next.next;
        return head;
    }
}

解法三：双指针

• 定义两个指针，一个first，一个second。
• first指针先走n步
• 再将first和second两个指针，同时后移，直到first为null。
• 则second指向的是倒是n+1个节点。

代码为：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0,head);
        ListNode first = head;
        ListNode second = dummy;
        for (int i = 1;i<=n;i++){
            first = first.next;
        }
        while (first!=null){
            first = first.next;
            second = second.next;
        }
        second.next = second.next.next;
        ListNode node = dummy.next;
        return node;
    }
}

不用辅助节点时：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode first = head;
        ListNode second = head;
        for (int i = 1;i<=n;i++){
            first = first.next;//使得first与second之间间隔n-1个节点
        }
        //判断是否删除的是头结点
        if (first==null) return head.next;
        while (first.next!=null){//将index2移至最后一个节点	
            first = first.next;
            second = second.next;
        }
        second.next = second.next.next;
        return head;
    }
}