19. Remove Nth Node From End of List

题目

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.

分析

题目给了一个链表的一维数组，和一个整数n，要求删除从末尾开始数第n个节点，然后返回链表的头。做法是，用两个指针，一个fast指针，一个solw指针，让slow指针永远慢fast指针n个节点。fast走完后，再用slow指针来删除需要删除的节点

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) {
        ListNode* h = new ListNode(0);
        h->next = head;
        ListNode* slow = h;
        ListNode* fast = h;
        while(fast->next!=NULL){
            if(n<=0){
                slow = slow->next;
            }
            fast= fast->next;
            n--;
        }
        if(slow->next!=NULL){
            slow->next = slow->next->next;
        }
        return h->next;
    }
};