Remove Nth Node From End of List

Problem

Given a linked list, remove the nth node from the end of list and return its head.

Notice

The minimum number of nodes in list is n.

Example

Given linked list: 1->2->3->4->5->null, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5->null.

Challenge

Can you do it without getting the length of the linked list?

题解

简单题,使用快慢指针解决此题,需要注意最后删除的是否为头节点。让快指针先走n步,直至快指针走到终点,找到需要删除节点之前的一个节点,改变node->next域即可。见基础数据结构部分的链表解析。

C++

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/**
* Definition of ListNode
* class ListNode {
* public:
* int val;
* ListNode *next;
* ListNode(int val) {
* this->val = val;
* this->next = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param head: The first node of linked list.
* @param n: An integer.
* @return: The head of linked list.
*/
ListNode *removeNthFromEnd(ListNode *head, int n) {
if (NULL == head || n < 1) {
return head;
}

ListNode dummy(0);
dummy.next = head;
ListNode *preDel = dummy;

for (int i = 0; i != n; ++i) {
if (NULL == head) {
return NULL;
}
head = head->next;
}

while (head) {
head = head->next;
preDel = preDel->next;
}
preDel->next = preDel->next->next;

return dummy.next;
}
};

Java

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/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
if (head == nul) return head;

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

while(fast != null) {
fast = fast.next;
slow = slow.next;
}

// gc friendly
// ListNode toBeDeleted = slow.next;
slow.next = slow.next.next;
// toBeDeleted.next = null;
// toBeDeleted = null;

return dummy.next;
}
}

源码分析

引入dummy节点后画个图分析下就能确定headpreDel的转移关系了。 注意 while 循环中和快慢指针初始化的关系,否则容易在顺序上错一。

复杂度分析

极限情况下遍历两遍链表,时间复杂度为 $$O(n)$$.


   链表


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