[LeetCode] Remove Nth Node From End of List

最新推荐文章于 2024-01-12 15:01:11 发布

最新推荐文章于 2024-01-12 15:01:11 发布 · 78 阅读

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本文介绍如何在单链表中删除倒数第N个节点，并提供了一个有效的算法实现，确保操作在一次遍历中完成。

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.

/**
 * 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) {
        if (head == NULL) return head;
        ListNode* first = head, *second = head;
        while (n > 1 && first != NULL) first = first->next, n--;
        if (n > 1 || first == NULL) return head;
        if (first->next == NULL) {
            ListNode* newhead = head->next;
            delete head;
            return newhead;
        }
        ListNode* prev;
        while (first->next != NULL) {
            prev = second;
            first = first->next;
            second = second->next;
        }
        prev->next = second->next;
        delete second;
        return head;
    }
};