【Leetcode长征系列】Remove Nth Node From End of List

原题：

Given a linked list, remove the n^th 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.

思路：用两个指针，让它们相隔n，当后一个指针指到最尾端时，前一个刚好指到我们需要删除的点。不过要一个额外的指针记录删除点的前一个节点。还需要考虑到删除头结点和末尾节点的情况。

/** * 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 *mark = NULL;        ListNode *tmp = head;        ListNode *del = head;                if (!head||!head->next) return NULL;        for(int i=0; i<n-1; i++)            tmp = tmp->next;                while(tmp->next){            mark = del;            del = del->next;            tmp = tmp->next;        }        if (mark==NULL){            head=del->next;            delete del;        }        else{            mark->next = del->next;            delete del;        }        return head;    }};