Leetcode:remove Nth Node from End of List

Given A linked list, remove the nth node from the end of the list and return its head.

For example,

   1->2->3->4->5  N = 2.   1->2->3->5.

Note:
Given n would always be valid.

Try to do the in one pass.

Algorithm: 1 prepare two pointers first, Second2 let FISRT walk n step 3 let fisrt and second go at the same time until you meet the end 4 to use a temp pointer to record second the previous node, used to delete second.

/** * Definition for singly-linked list. * struct ListNode {*     int val; *     ListNode *next; *     listnode (int x): Val (x), Next (NULL) {} *}; */class Soluti On {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;    };

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Leetcode:remove Nth Node from End of List