19. Remove Nth Node From End of List

问题描述

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

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.

Follow up:
Could you do this in one pass?
题目链接：

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思路分析

给一条链表，删除链表从尾部往前数第n个节点，要求只遍历一遍链表。

用两指针，从新的头begin出发，fast指针先从begin向前走n+1个节点，然后slow再出发，当fast到null了之后，slow的下一个节点就是要删除的了，删掉就行。

代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if (!head || !head->next)
            return NULL;
        ListNode* slow = head;
        ListNode* fast = head;
        ListNode* begin = head;
        while (fast->next && fast->next->next){
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast){
                while (begin != slow){
                    begin = begin->next;
                    slow = slow->next;
                }
                return begin;
            }
        }
        return NULL;
    }
};

时间复杂度：$O(n)$
空间复杂度：$O(1)$

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反思

快慢指针是处理链表中环的不二选择+1，还要考虑好头的问题，如果从head出发，则有可能head无法删除。