Remove Nth Node From End of List

描述

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

For example, 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.

Note:

• Given n will always be valid.
• Try to do this in one pass.

分析

设两个指针p,q，让q先走n步，然后p和q一起走，直到q走到尾节点，删除p->next即可。

代码

Java

// Remove Nth Node From End of List
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode p = dummy, q = dummy;

        for (int i = 0; i < n; i++)  // q先走n步
            q = q.next;

        while(q.next != null) { // 一起走
            p = p.next;
            q = q.next;
        }
        p.next = p.next.next;
        return dummy.next;
    }
}

C++

// Remove Nth Node From End of List
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode dummy{-1, head};
        ListNode *p = &dummy, *q = &dummy;

        for (int i = 0; i < n; i++)  // q先走n步
            q = q->next;

        while(q->next != nullptr) { // 一起走
            p = p->next;
            q = q->next;
        }
        ListNode *tmp = p->next;
        p->next = p->next->next;
        delete tmp;
        return dummy.next;
    }
};

Python

# Remove Nth Node From End of List
# 时间复杂度O(n)，空间复杂度O(1)
class Solution:
    def removeNthFromEnd(self, head, n):
        dummy = ListNode(-1)
        dummy.next = head
        p, q = dummy, dummy

        for i in range(n): # q先走n步
            q = q.next

        while q.next != None: # 一起走
            p = p.next
            q = q.next
        p.next = p.next.next
        return dummy.next

相关题目

• Intersection of Two Linked Lists