[LeetCode] Remove Nth Node From End of List

最新推荐文章于 2024-01-12 15:01:11 发布

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本文介绍了一种高效算法，用于从单链表中移除倒数第N个节点，仅通过一次遍历即可完成操作。该算法利用两个指针之间的固定距离特性来定位目标节点，并提供了一个示例代码实现。

Given a linked list, remove the nth 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.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode *p = head, *q = head, *pre = nullptr;
        for(int i = 0; i < n-1; i++)    q = q->next;
        while(q->next)
        {
            pre = p;
            p = p->next;
            q = q->next;
        }
        if(pre == nullptr)
        {
            head = p->next;
            delete p;
        }
        else
        {
            pre->next = p->next;
            delete p;
        }
        return head;
    }
};

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode dummy(0);
        dummy.next = head;
        ListNode *p = &dummy, *q = &dummy;
        while(n--)  q = q->next;
        while(q->next)
        {
            p = p->next;
            q = q->next;
        }
        ListNode *tmp = p->next;
        p->next = tmp->next;
        delete tmp;
        return dummy.next;
    }
};