LeetCode--remove-nth-node-from-end-of-list

Miles-

于 2018-01-01 20:50:10 发布

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分类专栏： LeetCode 文章标签： LeetCode C++

本文链接：https://blog.youkuaiyun.com/wl_ss/article/details/78947287

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本文介绍了一种使用双指针技术在单次遍历中删除链表倒数第N个节点的方法。通过引入虚拟头节点简化边界条件处理，确保了算法的简洁性和有效性。

题目描述

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：利用双指针找到第n个数，然后删除之。

/**
 * 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 == nullptr)
            return head;
        
        ListNode* fake = new ListNode(0);
        fake->next = head;
        ListNode* slow = fake;
        ListNode* fast = fake;
        
        for(int i=0; i<n; ++i)
            fast = fast->next;
        
        while(fast->next != nullptr)
        {
            fast = fast->next;
            slow = slow->next;
        }
        
        ListNode* temp = slow->next;
        slow->next = slow->next->next;
        delete temp;
        return fake->next;
    }
};