19. 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.

思考：一道非常基础的题目。找到要删除的节点的前一个节点，以上题为例，欲删除节点为4，那么找到节点3。怎么找到节点3呢?用两个指针相隔n+1，然后一起往前走，直到最后。代码实现如下。

/**
 * 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 *p = head;
        ListNode *pre = NULL;
        int count = 0;
        while(p) { 
            count++;
            p = p->next;
            if(count==n+1) {
                pre = head; 
            }else if(count>n+1) {
                pre = pre->next;
            }
        }
        
        if(count<n) return NULL;
        if(count==n) return head->next;
        //count>n means the link have more than n node.
        pre->next = pre->next->next;
        
        return head;
    }
};