19. Remove Nth Node From End of List

• 第一种解法

  /**
   * 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) 
  	{
  		int count = 0;
  		ListNode* pre = head;  
  		if(head == NULL)
  		{
  			return NULL;
  		}
  		
          while(pre != NULL)
  		{
  			count++;
  			pre = pre->next;
              
  		}
          
  		if(count == n)  //删除头节点
  		{
  			pre = head;
  		    return pre->next;
  		}
  		
  		pre = head;
          int i = 1;
  		while(pre != NULL)
  		{
  			if(i == count - n)
  			{
  				break;
  			}
  			i++;
              pre = pre->next;
  		}
  		//删除一个节点
  		ListNode* temp;
  	    temp = pre->next;
          
  		pre->next = temp->next;
  		delete temp;
  		
  		return head;
      }
  };

链表删除节点关键是找到待删除节点的前一个节点。另外就是注意要删除头节点的情况。在上面的代码中

  		if(count == n)  //删除头节点
  		{
  			pre = head;
  		    return pre->next;
  		}

只是把头节点的下一个元素返回了，并没有真正意义上的释放头节点指向的内存。因此将代码修改如下：

• 修改后代码

    /**
     * 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) 
    	{
    		int count = 0;
    		ListNode* pre = head;  
    		if(head == NULL)
    		{
    			return NULL;
    		}
    		
            while(pre != NULL)
    		{
    			count++;
    			pre = pre->next;
    		}
            
    		if(count == n)  //删除头节点
    		{
                pre = head;
                head = head->next;
                delete pre;
                return head;
    		}
    		
    		pre = head;
             int i = 1;
    		while(pre != NULL)
    		{
    			if(i == count - n)
    			{
    				break;
    			}
    			i++;
                pre = pre->next;
    		}
    		//删除一个节点
    		ListNode* temp;
    	    temp = pre->next;
            
    		pre->next = temp->next;
    		delete temp;
    		
    		return head;
        }
    };

• 快慢指针法

  声明一个快指针fast和一个慢指针slow。首先fast移动到待删除节点temp之前，然后fast和slow每次移动一步，当fast移动到最后时，slow指针正好移动到待删除节点temp之前，这时完成删除操作即可。

  /**
   * 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 == NULL)
  		{
  			return NULL;
  		}
  		
  		ListNode* newhead = new ListNode(0);
  		newhead->next = head;
  		
          ListNode* fast = newhead;
  		ListNode* slow = newhead;
  		
  		for(int i = 0;i < n;i++)
  		{
  			fast = fast->next;  
  		}
  		
  		while(fast->next)
  		{
  			fast = fast->next;
  			slow = slow->next;
  		}
  		
  		ListNode* temp = slow->next;  //temp表示待删除的节点
          slow->next = slow->next->next;
  				
  		delete temp;
  		
  		return newhead->next;
      }
  };