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.

/**
 * 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 || n < 1 ) return head;   //若head为空，或n不合法
        ListNode node(0);   //设置头结点，好删除
        node.next = head;
        ListNode* pre = &node;  //slow的前驱节点
        ListNode* fast = head;
        while( n && fast ) {    //令fast先指向第n个节点
            fast = fast->next;
            --n;
        }
        if( n && !fast ) return head;   //若链表长度没有n大
        ListNode* slow = head;
        while( fast ) { //fast和slow同时后移，slow为空时结束
            fast = fast->next;
            pre = slow;
            slow = slow->next;
        }
        pre->next = slow->next; //删除第n个节点
        delete slow;
        return node.next;
    }
};

 

设立前后两指针，假设为前指针为fast，后指针为slow，先让fast指针指向head，向后走n步，指向第n个节点，然后slow指针指向head，让fast指针和slow指针同时后移，知道slow为空，那么fast指针指向的就是倒数第n个节点