LeetCode 19. Remove Nth Node From End of List

用prev, slow, fast三个指针实现。

prev为slow的前一个指针，fast为slow的后n个指针。

当fast为NULL时，slow即为倒数第n个，此时将其删去即可；我的程序需要特判prev为NULL, 即slow为head的情况。

代码：

class Solution {public:    ListNode *removeNthFromEnd(ListNode *head, int n)     {    ListNode* slow = head, *fast = head, *prev = NULL;    for (int i = 0; i<n; ++ i, fast=fast->next) {}    while (fast != NULL){    prev = slow;slow = slow->next;fast = fast->next;}// note that we didn't delete the n-th node hereif (prev != NULL){    prev->next = slow->next;} else{    head = head->next;}return head;    }};