[LeetCode]19. Remove Nth Node From End of List

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.

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// 设置一个哑节点，两个指针，其中一个先行n步/** * 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* dummyHead = new ListNode(-1);        dummyHead->next = head;        ListNode* fast = dummyHead;        ListNode* slow = dummyHead;        while(n--){            if(fast)                fast = fast->next;            else                return NULL;        }        while(fast && fast->next != NULL){            fast = fast->next;            slow = slow->next;        }        ListNode* delNode = slow->next;        slow->next = slow->next->next;        delete delNode;        return dummyHead->next;    }};