#Leetcode# 19. Remove Nth Node From End of List

https://leetcode.com/problems/remove-nth-node-from-end-of-list/

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:

Could you do this in one pass?

代码：

/**
 * 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 || !head -> next) return NULL;
        ListNode *slow = head;
        ListNode *fast = head;
        
        for(int i = 1; i <= n; i ++) fast = fast -> next;
        if(!fast) return head -> next;
        while(fast -> next) {
            slow = slow -> next;
            fast = fast -> next;
        }
        
        slow -> next = slow -> next -> next;
        return head;
    }
};

　　仿佛写的还算顺利