removeNthFromEnd

# *- coding:utf-8   -*# 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):        """        :type head: ListNode        :type n: int        :rtype: ListNode        """        fast=slow=head        for x in range(n):        fast=fast.next        if not fast:        return head.next        while fast.next:        fast=fast.next        slow=slow.next        slow.next=slow.next.next        return head
C代码：
/*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.删除链表从后面数第n个节点e，需要知道链表从后面数第n+1个节点e_r的指针若链表总结点数为sum,则e_r的在链表中的位置sum-n (从链表头开始数)采用前后指针法来找到e_r的位置，只需遍历一遍链表,前后指针相差的值就是n,当前指针从链表头遍历到尾部时，后指针正好遍历到e_r的位置。*//** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* removeNthFromEnd(struct ListNode* head, int n){    struct ListNode* front=head;    struct ListNode* back=head;struct ListNode* temp=NULL;while(front!=NULL){front=front->next;if(n-- < 0 ) back=back->next;}if(n==0) {temp=head;head=head->next;}else{temp=back->next;back->next=temp->next;}free(temp);return head;     }