leetcode--19. 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.

我的解法好垃圾

我先确定表长，再用表长减去n，就得到正向要走多少次。

具体细节：先定义fast和slow两个指针，fast一次走2步，slow一次走1部，再定义一个size计数。当fast走到底时，size就是表的一半长度，(flag的目的是为了判断fast能否走2步，这很重要，确定表的长度是奇数还是偶数).确定了表长后，就是正常的链表遍历+删除了

/**
 * 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 *fast=head;
        ListNode *slow=head;
        int size=0;
        bool flag=false;
        if(head->next==nullptr)
        {
            delete head;
            return NULL;
        }
        while(fast->next!=NULL){
            fast=fast->next;
            slow=slow->next;
            flag=false;
            size++;
            if(fast->next!=NULL){
                fast=fast->next;
                flag=true;
            }
        }
        if(flag)
            size=size*2+1;
        else
            size=size*2;
        int walk = size-n;
        if(walk==0)
        {
            slow=head;
            head=head->next;
            delete slow;
            return head;
        }
        else
        {
            slow=head;
            while(walk!=1){
                slow = slow->next;
                walk--;
            }
            fast=slow->next;
            slow->next=fast->next;
            delete fast;
            return head;           
        }

    }
};

在ac后，看了discuss里的做法.先放一个大神解法

思路: t2先走n步，才让t1,t2同时走，直到t2走到头。这样，t1走的步数即使正确的了。

还有一点，t1使用二重指针，这样可以优雅的删除节点。

/**
 * 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** 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;
    }
};