leetcode: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.

算法：

1 准备两个指针first, second

2 先让fisrt走n步

3 让fisrt和second同一时候走直到first遇到结尾

4 要用一个temp指针来记录second前一个node。用来删除second用。

/**
 * 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 *first = head, *second = head;
     ListNode *temp = second;
 
     for(int i = 0; i < n; i++) {
         first = first -> next;
     }
 
     while(first) {
         first = first -> next;
         temp = second;
        second = second -> next;
     }
 
     if(second == head) {
         head = head -> next;
     }
     else {
         temp -> next = second -> next;
     }
 
 
     return head;
    }
};