【Leetcode】【Easy】Remove Nth Node From End of List

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.

 

普通青年解法：

设置三个指针A\B\C，指针A和B间隔n-1个结点，B在前A在后，用指针A去遍历链表直到指向最后一个元素，则此时指针B指向的结点为要删除的结点，指针C指向指针B的pre，方便删除的操作。

代码：

/**
 * 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 *target = head;
        ListNode *target_pre = NULL;
        ListNode *findthelast = head;
        
        if (head == NULL)
            return head;
        
        while(--n>0)
            findthelast = findthelast->next;
        
        while (findthelast->next) {
            target_pre = target;
            target = target->next;
            findthelast = findthelast->next;
        }
        
        if (target_pre == NULL) {
            head = target->next;
            return head;
        }
        
        target_pre->next = target->next;
        
        return head;
    }
};

 

文艺智慧青年解法：

 （需要更多理解）

class Solution
{
public:
    ListNode* removeNthFromEnd(ListNode* head, int n)
    {
        ListNode** t1 = &head, *t2 = head;
        for(int i = 1; i < n; ++i)
        {
            t2 = t2->next;
        }
        while(t2->next != NULL)
        {
            t1 = &((*t1)->next);
            t2 = t2->next;
        }
        *t1 = (*t1)->next;
        return head;
    }
};

 

附录：

C++二重指针深入