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。
