学习完单链表,习题就成了最好的巩固方式
目录
1.链表分割:
链表分割,链接奉上
思路:
我们可以创建两个newhead
,
将比x
小的尾插放到newhead1
中
比x
大的放在newhead2
中
再将两个链表进行链接
注意:
-
在进行创建新的链表时,最好使用带有哨兵位 的链表,在进行链接时会比较容易,
因为没有哨兵位的链表需要判断是否两个链表是否为空,并分别做出处理措施
-
在尾插时,有可能出现如下图的情况
故我们需要将
tail2->next = NULL
代码实现:
c
class Partition {
public:
ListNode* partition(ListNode* pHead, int val) {
//两个哨兵位的创建
struct ListNode* newhead1 = (struct ListNode*)malloc(sizeof(struct ListNode));
struct ListNode* newhead2 = (struct ListNode*)malloc(sizeof(struct ListNode));
struct ListNode* tail1 = newhead1;
struct ListNode* tail2 = newhead2;
tail1->next = newhead1->next = NULL;
tail2->next = newhead2->next = NULL;
while(pHead)
{
if(pHead->val < val)
{
tail1->next = pHead;
tail1 = tail1->next;
}
else
{
tail2->next = pHead;
tail2 = tail2->next;
}
pHead = pHead->next;
}
tail1->next = newhead2->next;
tail2->next = NULL;
return newhead1->next;
}
};
2.随机链表的复制:
思路1:
先复制一份没有进行random
处理的copy
链表
我们可以根据random的指向的位置 与原位置 ,
通过计算得出他们之间的距离,
再根据距离进行确定copy
链表中random
的位置
这样实现,时间复杂度是O(N^2)
代码实现:
理论存在,实践开始
c
/**
* Definition for a Node.
* struct Node {
* int val;
* struct Node *next;
* struct Node *random;
* };
*/
struct Node* creat(int x)
{
struct Node* newnode = (struct Node*)malloc(sizeof(struct Node));
newnode->val = x;
newnode->next = NULL;
return newnode;
}
struct Node* copyRandomList(struct Node* head)
{
struct Node* cur = head;
struct Node* newhead = NULL;
struct Node* tail = NULL;
while(cur)
{
if(newhead == NULL)
{
newhead = creat(cur->val);
tail = newhead;
}
else
{
tail->next = creat(cur->val);
tail = tail->next;
}
cur = cur->next;
}
int count = 0;
cur = head;
struct Node* subcur = newhead;
while(cur)
{
count = 0;
struct Node* tmp1 = head;
while(cur->random != tmp1)
{
count++;
tmp1 = tmp1->next;
}
struct Node* tmp = newhead;
while(count--)
{
tmp = tmp->next;
}
subcur->random = tmp;
cur = cur->next;
subcur = subcur->next;
}
return newhead;
}
思路2:
在原链表的每一个元素后边插入一个一样元素的copy节点
之后,我们发现当前节点后的copy节点的random指向当前节点的random的next ,仔细咀嚼(可以自己画图感受一下),这句话也是整个代码的核心
最后再将整个链表进行拆解,返回新的结点
代码实现:
c
/**
* Definition for a Node.
* struct Node {
* int val;
* struct Node *next;
* struct Node *random;
* };
*/
struct Node* copyRandomList(struct Node* head)
{
struct Node* cur = head;
//添加copy节点
while(cur)
{
struct Node* copy = (struct Node*)malloc(sizeof(struct Node));
struct Node* next = cur->next;
copy->val = cur->val;
copy->next = next;
cur->next = copy;
cur = next;
}
cur = head;
//random的复制
while(cur)
{
if(cur->random == NULL)
{
cur->next->random = NULL;
}
else
{
cur->next->random = cur->random->next;
}
cur = cur->next->next;
}
//解开节点
struct Node* newhead = NULL;
struct Node* tail = NULL;
cur = head;
while(cur)
{
struct Node* next = cur->next->next;
if(newhead == NULL)
{
newhead = tail = cur->next;
}
else
{
tail->next = cur->next;
tail = tail->next;
}
cur = next;
}
return newhead;
}
3.环形链表:
3.1环形链表1:
思路:
利用快慢指针
,这是解决此问题的最可行办法,
一个走一步,一个走两步,
当有环时,两者都进入环,因为两者固定减少1步,故必然可以相遇,
当没有环时,fast == NULL,或者fast指针->next == NULL
代码实现:
c
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
bool hasCycle(struct ListNode *head)
{
struct ListNode* slow = head;
struct ListNode* fast = head;
while(fast && fast->next)
{
slow = slow->next;
fast = fast->next->next;
if(slow == fast)
return true;
}
return false;
}
3.2环形链表2:
思路:
先输出一个结论:
找到快慢指针相遇的点,设置两个指针,一个从头开始走,一个从相遇点开始走,一次一步,相遇点就是环的开始点,我们返回此指针
代码实现:
c
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode *detectCycle(struct ListNode *head)
{
struct ListNode* slow = head;
struct ListNode* fast = head;
struct ListNode* cur = head;
while(fast && fast->next)
{
slow = slow->next;
fast = fast->next->next;
if(slow == fast)
{
struct ListNode* meet = slow;
while(1)
{
if(cur == meet)
{
return meet;
}
cur = cur->next;
meet = meet->next;
}
}
}
return NULL;
}
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