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0019.删除链表的倒数第N个节点.md

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参与本项目，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！

19.删除链表的倒数第N个节点

力扣题目链接

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。

进阶：你能尝试使用一趟扫描实现吗？

示例 1：

19.删除链表的倒数第N个节点

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]

思路

《代码随想录》算法公开课：链表遍历学清楚！ | LeetCode：19.删除链表倒数第N个节点，相信结合视频在看本篇题解，更有助于大家对链表的理解。

双指针的经典应用，如果要删除倒数第n个节点，让fast移动n步，然后让fast和slow同时移动，直到fast指向链表末尾。删掉slow所指向的节点就可以了。

思路是这样的，但要注意一些细节。

分为如下几步：

  • 首先这里我推荐大家使用虚拟头结点，这样方便处理删除实际头结点的逻辑，如果虚拟头结点不清楚，可以看这篇： 链表：听说用虚拟头节点会方便很多？

  • 定义fast指针和slow指针，初始值为虚拟头结点，如图：

  • fast首先走n + 1步 ，为什么是n+1呢，因为只有这样同时移动的时候slow才能指向删除节点的上一个节点（方便做删除操作），如图：

  • fast和slow同时移动，直到fast指向末尾，如题：

  • 删除slow指向的下一个节点，如图：

此时不难写出如下C++代码：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* dummyHead = new ListNode(0);
        dummyHead->next = head;
        ListNode* slow = dummyHead;
        ListNode* fast = dummyHead;
        while(n-- && fast != NULL) {
            fast = fast->next;
        }
        fast = fast->next; // fast再提前走一步，因为需要让slow指向删除节点的上一个节点
        while (fast != NULL) {
            fast = fast->next;
            slow = slow->next;
        }
        slow->next = slow->next->next;
        return dummyHead->next;
    }
};

其他语言版本

java:

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(-1);
        dummy.next = head;

        ListNode slow = dummy;
        ListNode fast = dummy;
        while (n-- > 0) {
            fast = fast.next;
        }
        // 记住 待删除节点slow 的上一节点
        ListNode prev = null;
        while (fast != null) {
            prev = slow;
            slow = slow.next;
            fast = fast.next;
        }
        // 上一节点的next指针绕过 待删除节点slow 直接指向slow的下一节点
        prev.next = slow.next;
        // 释放 待删除节点slow 的next指针, 这句删掉也能AC
        slow.next = null;

        return dummy.next;
    }
}

Python:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        head_dummy = ListNode()
        head_dummy.next = head

        slow, fast = head_dummy, head_dummy
        while(n!=0): #fast先往前走n步
            fast = fast.next
            n -= 1
        while(fast.next!=None):
            slow = slow.next
            fast = fast.next
        #fast 走到结尾后，slow的下一个节点为倒数第N个节点
        slow.next = slow.next.next #删除
        return head_dummy.next

Go:

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func removeNthFromEnd(head *ListNode, n int) *ListNode {
    dummyHead := &ListNode{}
    dummyHead.Next = head
    cur := head
    prev := dummyHead
    i := 1
    for cur != nil {
        cur = cur.Next
        if i > n {
            prev = prev.Next
        }
        i++
    }
    prev.Next = prev.Next.Next
    return dummyHead.Next
}

JavaScript:

/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd = function(head, n) {
    let ret = new ListNode(0, head),
        slow = fast = ret;
    while(n--) fast = fast.next;
    while (fast.next !== null) {
        fast = fast.next; 
        slow = slow.next
    };
    slow.next = slow.next.next;
    return ret.next;
};

TypeScript:

版本一（快慢指针法）：

function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
    let newHead: ListNode | null = new ListNode(0, head);
    let slowNode: ListNode | null = newHead,
        fastNode: ListNode | null = newHead;
    for (let i = 0; i < n; i++) {
        fastNode = fastNode.next;
    }
    while (fastNode.next) {
        fastNode = fastNode.next;
        slowNode = slowNode.next;
    }
    slowNode.next = slowNode.next.next;
    return newHead.next;
};

版本二（计算节点总数法）：

function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
    let curNode: ListNode | null = head;
    let listSize: number = 0;
    while (curNode) {
        curNode = curNode.next;
        listSize++;
    }
    if (listSize === n) {
        head = head.next;
    } else {
        curNode = head;
        for (let i = 0; i < listSize - n - 1; i++) {
            curNode = curNode.next;
        }
        curNode.next = curNode.next.next;
    }
    return head;
};

版本三（递归倒退n法）：

function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
    let newHead: ListNode | null = new ListNode(0, head);
    let cnt = 0;
    function recur(node) {
        if (node === null) return;
        recur(node.next);
        cnt++;
        if (cnt === n + 1) {
            node.next = node.next.next;
        }
    }
    recur(newHead);
    return newHead.next;
};

Kotlin:

fun removeNthFromEnd(head: ListNode?, n: Int): ListNode? {
    val pre = ListNode(0).apply {
        this.next = head
    }
    var fastNode: ListNode? = pre
    var slowNode: ListNode? = pre
    for (i in 0..n) {
        fastNode = fastNode?.next
    }
    while (fastNode != null) {
        slowNode = slowNode?.next
        fastNode = fastNode.next
    }
    slowNode?.next = slowNode?.next?.next
    return pre.next
}

Swift：

func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? {
    if head == nil {
        return nil
    }
    if n == 0 {
        return head
    }
    let dummyHead = ListNode(-1, head)
    var fast: ListNode? = dummyHead
    var slow: ListNode? = dummyHead
    // fast 前移 n
    for _ in 0 ..< n {
        fast = fast?.next
    }
    while fast?.next != nil {
        fast = fast?.next
        slow = slow?.next
    }
    slow?.next = slow?.next?.next
    return dummyHead.next
}

PHP:

function removeNthFromEnd($head, $n) {
    // 设置虚拟头节点
    $dummyHead = new ListNode();
    $dummyHead->next = $head;

    $slow = $fast = $dummyHead;
    while($n-- && $fast != null){
        $fast = $fast->next;
    }
    // fast 再走一步，让 slow 指向删除节点的上一个节点
    $fast = $fast->next; 
    while ($fast != NULL) {
        $fast = $fast->next;
        $slow = $slow->next;
    }
    $slow->next = $slow->next->next;
    return $dummyHead->next;
    }

Scala:

object Solution {
  def removeNthFromEnd(head: ListNode, n: Int): ListNode = {
    val dummy = new ListNode(-1, head) // 定义虚拟头节点
    var fast = head // 快指针从头开始走
    var slow = dummy // 慢指针从虚拟头开始头
    // 因为参数 n 是不可变量，所以不能使用 while(n>0){n-=1}的方式
    for (i <- 0 until n) {
      fast = fast.next
    }
    // 快指针和满指针一起走，直到fast走到null
    while (fast != null) {
      slow = slow.next
      fast = fast.next
    }
    // 删除slow的下一个节点 
    slow.next = slow.next.next
    // 返回虚拟头节点的下一个
    dummy.next
  }
}