LeetCode刷題筆記（鏈表）：remove-nth-node-from-end-of-list

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• 轉載請註明作者和出處：http://blog.csdn.net/u011475210
• 代碼地址：https://github.com/WordZzzz/Note/tree/master/LeetCode
• 刷題平臺：https://www.nowcoder.com/ta/leetcode
• 題  庫：Leetcode經典編程題
• 編  者：WordZzzz

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題目描述

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.

解題思路

刪除倒數第n個結點，我們就需要找到倒數第n個結點。於是，我們可以利用快慢指針來實現。

首先，fast走n步指向第n個結點；接着，fast和slow一起走，直到fast指向尾結點；最後，刪除元素（這裏需要注意的是加入pre來判斷刪除元素是否爲頭結點）。

C++版代碼實現

/**
 * 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 NULL;

        ListNode *slow = head;
        ListNode *fast = head;
        ListNode *pre = NULL;
        //fast先走n步，到達第n個結點
        while(--n)
            fast = fast->next;
        //fast和slow一起走，直到fast走到鏈表尾部
        while(fast->next != NULL){
            pre = slow;
            slow = slow->next;
            fast = fast->next;
        }
        //此處用於判斷刪除的結點是否爲頭結點
        if(pre != NULL)
            pre->next = slow->next;
        else
            head = head->next;

        return head;
    }
};

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