21. Merge Two Sorted Lists

• 出错误的解法一

  /**
   * Definition for singly-linked list.
   * struct ListNode {
   *     int val;
   *     ListNode *next;
   *     ListNode(int x) : val(x), next(NULL) {}
   * };
   */
  class Solution
  {
  public:
      ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) 
  	{
          ListNode* newnode = new ListNode(0);
  		newnode->next = NULL;
  		ListNode* pre = newnode;
  		
  		if(l1 == NULL && l2 == NULL)
  		{
  			return NULL;
  		}
  		
  		if(l1 == NULL && l2 != NULL)
  		{
  			return l2;
  		}
  		
  		if(l2 == NULL && l1 != NULL)
  		{
  			return l1;
  		}
  		
  		while(l1 != NULL && l2 != NULL)
  		{
  			if(l1->val == l2->val)
  			{
  				//l1再前 l2再后 l1 l2均插入新队列
  				pre->val = l1->val;
  				pre->next = new ListNode(0);
  				pre->val = l2->val;
  				pre->next = NULL;
  				l1 = l1->next;
  				l2 = l2->next;
  			}
  			else if(l1->val > l2->val)
  			{
  				//l2插入新队列
  				pre->val = l2->val;
  				pre->next = new ListNode(0);
  				l2 = l2->next;
  			}
  			else
  			{
  				//l1插入新队列
  				pre->val = l1->val;
  				pre->next = new ListNode(0);
  				l1 = l1->next;
  			}
  			
  			
  		}
  		
  		while(l1 != NULL)
  		{
  			pre->next = new ListNode(0);
  			pre->val = l1->val;
  			l1 = l1->next;
  		}
  		
  		while(l2 != NULL)
  		{
  			pre->next = new ListNode(0);
  			pre->val = l2->val;			
  			l2 = l2->next;
  		}
  		
  		return newnode;
      } 
  };

• 出错误的解法二

  class Solution
  {
  public:
  	ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
  	{
  		ListNode* pre = NULL;
  		ListNode* newnode = pre;  //尾指针初始状态下指向头节点
  
  		if (l1 == NULL && l2 == NULL)
  		{
  			return NULL;
  		}
  
  		if (l1 == NULL && l2 != NULL)
  		{
  			return l2;
  		}
  
  		if (l2 == NULL && l1 != NULL)
  		{
  			return l1;
  		}
  
  		while (l1 != NULL && l2 != NULL)
  		{
  			if (l1->val == l2->val)
  			{
  				pre = new ListNode(0);
  				pre->next = NULL;
  				pre->val = l1->val;
  				pre = pre->next;
  				l1 = l1->next;
  			}
  			else if (l1->val > l2->val)
  			{
  				//l2插入新队列
  				pre = new ListNode(0);
  				pre->next = NULL;
  				pre->val = l2->val;
  				pre = pre->next;
  				l2 = l2->next;
  			}
  			else
  			{
  				//l1插入新队列
  				pre = new ListNode(0);
  				pre->next = NULL;
  				pre->val = l1->val;
  				pre = pre->next;
  				l1 = l1->next;
  			}
  
  
  		}
  
  		while (l1 != NULL)
  		{
  			pre = new ListNode(0);
  			pre->next = NULL;
  			pre->val = l1->val;
  			pre = pre->next;
  			l1 = l1->next;
  		}
  
  		while (l2 != NULL)
  		{
  			pre = new ListNode(0);
  			pre->next = NULL;
  			pre->val = l2->val;
  			pre = pre->next;
  			l2 = l2->next;
  		}
  
  		return newnode;
  	}
  };

• AC版代码1

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution
{
public:
	ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
	{
		ListNode* newnode = new ListNode(0);  //头节点
		ListNode* pre = newnode; //尾指针初始状态下指向头节点

		if (l1 == NULL && l2 == NULL)
		{
			return NULL;
		}

		if (l1 == NULL && l2 != NULL)
		{
			return l2;
		}

		if (l2 == NULL && l1 != NULL)
		{
			return l1;
		}

		while (l1 != NULL && l2 != NULL)
		{
			if (l1->val == l2->val)
			{
				pre->next = new ListNode(0);
				pre = pre->next;
				pre->val = l1->val;
				l1 = l1->next;
			}
			else if (l1->val > l2->val)
			{
				//l2插入新队列
				pre->next = new ListNode(0);
				pre = pre->next;
				pre->val = l2->val;
				l2 = l2->next;
			}
			else
			{
				//l1插入新队列
				pre->next = new ListNode(0);
				pre = pre->next;
				pre->val = l1->val;
				l1 = l1->next;
			}


		}

		while (l1 != NULL)
		{
			pre->next = new ListNode(0);
			pre = pre->next;
			pre->val = l1->val;
			l1 = l1->next;
		}

		while (l2 != NULL)
		{
			pre->next = new ListNode(0);
			pre = pre->next;
			pre->val = l2->val;
			l2 = l2->next;
		}
		pre->next = NULL;
		return newnode->next;
	}
};

这一版代码其实有很大的优化空间，首先不需要一开始对 l1 和 l2 的判空，因为在while循环中已经处理过这个问题。经过优化后的代码如下:

• AC版代码2

    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    class Solution
    {
    public:
    	ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
    	{
    		ListNode* newnode = new ListNode(0);  //头节点
            newnode->next = NULL;
    		ListNode* pre = newnode; //尾指针初始状态下指向头节点
    
    		while (l1 != NULL && l2 != NULL)
    		{
    			if (l1->val == l2->val)
    			{
    				pre->next = new ListNode(0);
    				pre = pre->next;
    				pre->val = l1->val;
    				l1 = l1->next;
    			}
    			else if (l1->val > l2->val)
    			{
    				//l2插入新队列
    				pre->next = new ListNode(0);
    				pre = pre->next;
    				pre->val = l2->val;
    				l2 = l2->next;
    			}
    			else
    			{
    				//l1插入新队列
    				pre->next = new ListNode(0);
    				pre = pre->next;
    				pre->val = l1->val;
    				l1 = l1->next;
    			}
    
    
    		}
    
    		while (l1 != NULL)
    		{
    			pre->next = new ListNode(0);
    			pre = pre->next;
    			pre->val = l1->val;
    			l1 = l1->next;
    		}
    
    		while (l2 != NULL)
    		{
    			pre->next = new ListNode(0);
    			pre = pre->next;
    			pre->val = l2->val;
    			l2 = l2->next;
    		}
    		pre->next = NULL;
    		return newnode->next;
    	}
    };

• 另一种思路

这道题目并没有说不可以更改l1和l2，因此又有如下代码：

     /**
      * Definition for singly-linked list.
      * struct ListNode {
      *     int val;
      *     ListNode *next;
      *     ListNode(int x) : val(x), next(NULL) {}
      * };
      */
     class Solution
     {
     public:
     	ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
     	{
     		ListNode* headnode = new ListNode(0);
     		headnode->next = NULL;
     		ListNode* pre = headnode;  //尾指针初始状态下指向头节点
     		
     		//注意题目中l1和l2都没有头节点
     		while(l1 != NULL && l2 != NULL)
     		{
     			if(l1->val > l2->val)
     			{
     				pre->next = l2;
     				pre = pre->next;
     				l2 = l2->next;
     			}
     			else
     			{
     				//相等 或 l1 比 l2小 都应该执行这里
     				pre->next = l1;
     				pre = pre->next;
     				l1 = l1->next;
     			}
     			
     		}
     		
     		if(l1)
     		{
     			pre->next = l1;
     			l1 = NULL;
     		}
     		
     		if(l2)
     		{
     			pre->next = l2;
     			l2= NULL;
     		}
     		
     		return headnode->next;
     	}
     };

这次代码的思路是直接把l1或者 l2 的节点摘除贴在pre后面。