1. Remove Nth Node From End of List My Submissions QuestionEditorial Solution
   Total Accepted: 106592 Total Submissions: 361392 Difficulty: Easy
   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.

Submission Details

207 / 207 test cases passed.

Status: Accepted

Runtime: 8 ms

思路：利用快慢指针，快指针先走n步，然后快指针到末尾，那么慢指针走到第n个位置的前一个位置

/** * 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 *pre=head,*p=head;        for(int i=0;i
     
      next;                else return head;         }        if(p==NULL)return head->next; //第n+1个位置为空，那么倒数第n个位置为首元素        while(p->next){               pre = pre->next;   //快慢指针同时走，直到快指针到达最后一个节点               p = p->next;        }        pre->next = pre->next->next;//删除该节点        return head;    }};