LeetCode之“链表”：Remove Nth Node From End of List

　　题目链接

　　题目要求：

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

　　For example,

   Given linked list: ->->->->, and n = .

   After removing the second node from the end, the linked list becomes ->->->.

　　Note:

　　Given n will always be valid.
　　Try to do this in one pass.

　　这道题比较常规的想法是先算出链表长度len，再将头指针移动len-n步就可以到达待删节点了。具体程序如下（4ms）：

 /**
  * 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)
             return head;

         int len = ;
         ListNode *start = head;
         while(start)
         {
             len++;
             start = start->next;
         }

         if(len == n)
         {
             start = head;
             head = head->next;
             delete start;
             start = nullptr;
             return head;
         }

         ListNode *preNode = head;
         start = preNode->next;
         int k = ;
         while(k < len - n)
         {
             preNode = start;
             start = start->next;
             k++;
         }

         preNode->next = start->next;
         delete start;
         start = nullptr;

         return head;
     }
 };

　　另一个比较巧妙的方法就是先定义一个慢指针（初始值为头指针父指针），让头指针移动n步，再让头指针和一个慢指针同时移动直至头指针为空，这时慢指针指向的下一个节点就是待删节点。具体程序日如下（4ms）：

 /**
  * 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)
             return head;

         ListNode *dummy = new(nothrow) ListNode(INT_MIN);
         assert(dummy);
         dummy->next = head;

         for(int i = ; i < n; i++)
             head = head->next;

         ListNode *preNode = dummy;
         while(head)
         {
             head = head->next;
             preNode = preNode->next;
         }

         ListNode *delNode = preNode->next;
         preNode->next = delNode->next;
         delete delNode;
         delNode = nullptr;

         head = dummy->next;
         delete dummy;
         dummy = nullptr;

         return head;
     }
 };