leetcode_[python/C++]_19. Remove Nth Node From End of List(删除链表末第n个节点）

题目链接
【题目】
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.

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【分析】
做法（1）：一个指针+count len(链表）
做法（2）：两个指针
做法（3）：两个指针+增加一个pre指针
做法（4）：二级指针
做法（5）：递归
附上最优runtime

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C++:
一个指针+count len(链表)

/** * 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 * pre = head;        int count = 0;        while( pre != NULL ){            count ++;            pre = pre->next;        }        int m = count - n - 1;        int c = 0;        ListNode * p = head;        if( m == -1 ) return head->next;        while( c != m ){           c ++;           p = p->next;        }           ListNode *temp = p->next->next;        if(temp == NULL) p->next = NULL;        else p->next = temp;        return head;    }};

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两个指针+pre指针

class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        if( head == NULL ) return NULL;        ListNode * pre = NULL, * current = head, * end = head;        for(int i = 0 ; i < n ; i ++ ) end = end->next;        while( end ){            pre = current;            current = current->next;            end = end->next;        }        if( pre == NULL ) return head->next;        else pre->next = current->next;        return head;    }};

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链表头增加一个节点：

class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        if( head == NULL ) return NULL;        ListNode * new_head = new ListNode(0);        new_head->next = head;        ListNode * current = new_head, * end = new_head;        for(int i = 0 ; i <= n ; i ++ ) end = end->next;        while( end ){            current = current->next;            end = end->next;        }        current->next = current->next->next;        return new_head->next;    }};

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二级指针：
参考学习二级指针

class Solution{public:    ListNode* removeNthFromEnd(ListNode* head, int n)    {        ListNode** t1 = &head, *t2 = head;//t1 == pre   *t1 = current   (*t1)->next = current->next          for(int i = 0; i < n; ++i)        {            t2 = t2->next;        }        while(t2 != NULL)        {            t1 = &((*t1)->next);            t2 = t2->next;        }        *t1 = (*t1)->next;        return head;    }};

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递归：

class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int& n) {        ListNode * node = new ListNode(0);        node->next = head;        removeNode(node, n);        return node->next;    }    void removeNode(ListNode* head, int& n){        if(!head) return;        removeNode(head->next, n);        n--;        if(n == -1) {            head->next = head->next->next;         }        return;    }};

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python

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构造链表：

def removeNthFromEnd(self, head, n):    self.next, nodelist  = head, [self]    while head.next:        if len(nodelist) == n:            nodelist.pop(0)        nodelist += head,        head = head.next    nodelist[0].next = nodelist[0].next.next     return self.next

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新建链表：

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def removeNthFromEnd(self, head, n):        collection=list();        ptr=head;        collection.append(ptr)        while ptr.next is not None:            collection.append(ptr.next)            ptr=ptr.next        length=len(collection);        if length-n==0 and length>1:            return collection[1]        elif length-n==0 and length==1:            return None        elif n>1:            collection[length-n-1].next=collection[length-n+1]            return head        elif n==1:            collection[length-2].next=None            return head

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两个指针：

class Solution(object):    def removeNthFromEnd(self, head, n):        """        :type head: ListNode        :type n: int        :rtype: ListNode        """        current = end = self        self.next = head        while end.next:            if n:                n -= 1            else:                current = current.next            end = end.next        current.next = current.next.next        return self.next 

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两个指针另一种写法：

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def removeNthFromEnd(self, head, n):        current = head        end = head        for i in range(0,n):            end = end.next        if(end == None):            return head.next        while(end.next != None):            end = end.next            current = current.next        current.next = current.next.next        return head

这个效率比较高