[Leedcode] Necessary interview questions for linked lists in data structures (Phase 3)
Article directory
1. The first question
1. Topic
The CM11 linked list is divided as follows (example):
现有一链表的头指针ListNode*pHead，给一定值x，
编写一段代码将所有小于x的结点排在其余结点之前，且不能改变原来的数据顺序，返回重新排列后的链表的头指针。
2. Ideas
1. Use the head nodes lesshead and greathead with two sentinel positions , connect the ones less than 4 to the back of lesshead, and the ones greater than 4 to the back of greathead;
2. Then connect the last one less than 4 to the first one greater than 4: Specifically as shown below
Note: as shown below!
3. Source code
The code is as follows (example):
struct ListNode {
int val;
struct ListNode *next;
};
class Partition {
public:
ListNode* partition(ListNode* pHead, int x) {
struct ListNode* lessHead, *lessTail, *greaterHead, *greaterTail;
lessHead = lessTail = (struct ListNode*)malloc(sizeof(struct ListNode));
greaterHead = greaterTail = (struct ListNode*)malloc(sizeof(struct ListNode));
lessTail->next = greaterTail->next = NULL;
struct ListNode* cur = pHead;
while (cur) {
if (cur->val < x) {
lessTail->next = cur;
lessTail = lessTail->next;
}
else {
greaterTail->next = cur;
greaterTail = greaterTail->next;
}
cur = cur->next;
}
lessTail->next = greaterHead->next;
greaterTail->next = NULL;
struct ListNode* list = lessHead->next;
free(lessHead);
free(greaterHead);
return list;
}
};
2. The second question
1. Topic
The palindrome structure of OR36 linked list is as follows (example):
对于一个链表，请设计一个时间复杂度为O(n),额外空间复杂度为O(1)的算法，判断其是否为回文结构。
给定一个链表的头指针A，请返回一个bool值，代表其是否为回文结构。保证链表长度小于等于900。
2. Ideas
(1) The first case: an even number of linked lists
(2) The second case: an odd number of linked lists
3. Source code
(1) Implementation of the intermediate node of the linked list
- The implementation of the intermediate node of the linked list is as follows (example):
struct ListNode
{
int val;
struct ListNode *next;
}
struct ListNode* middleNode(struct ListNode* head)
{
struct ListNode* slow,*quick;
slow=quick=head;
while(quick && quick->next)
{
slow= slow->next;
quick= quick->next->next;
}
return slow;
}
(2) Implementation of reverse linked list
- The implementation of reverse linked list is as follows (example):
struct ListNode
{
int val;
struct ListNode *next;
};
struct ListNode* reverseList(struct ListNode* head)
{
//第二种方法
struct ListNode* newhead=NULL;
struct ListNode* cur=head;
while(cur)
{
struct ListNode* next=cur->next;
cur->next=newhead;
newhead=cur;
cur=next;
}
return newhead;
}
(3) Implementation of linked list comparison function
The code is as follows (example):
class PalindromeList {
public:
bool chkPalindrome(ListNode* A) {
struct ListNode* mid =middleNode(A);
struct ListNode* rHead =reverseList(mid);
struct ListNode* curA=A;
struct ListNode* curR = rHead;
while( curA && curR)
{
if(curA -> val != curR ->val)
{
return false;
}
else {
curA=curA->next;
curR =curR-> next;
}
}
return true;
}
};
(4) Overall source code
The code is as follows (example):
struct ListNode
{
int val;
struct ListNode *next;
ListNode(int x) : val(x), next(NULL) {
}
};
struct ListNode* middleNode(struct ListNode* head)
{
struct ListNode* slow,*quick;
slow=quick=head;
while(quick && quick->next)
{
slow= slow->next;
quick= quick->next->next;
}
return slow;
}
struct ListNode* reverseList(struct ListNode* head)
{
//第二种方法
struct ListNode* newhead=NULL;
struct ListNode* cur=head;
while(cur)
{
struct ListNode* next=cur->next;
cur->next=newhead;
newhead=cur;
cur=next;
}
return newhead;
}
class PalindromeList {
public:
bool chkPalindrome(ListNode* A) {
struct ListNode* mid =middleNode(A);
struct ListNode* rHead =reverseList(mid);
struct ListNode* curA=A;
struct ListNode* curR = rHead;
while( curA && curR)
{
if(curA -> val != curR ->val)
{
return false;
}
else {
curA=curA->next;
curR =curR-> next;
}
}
return true;
}
};
Summarize
The above is what I want to talk about today. This article introduces the necessary interview questions for linked lists in data structures (the third issue).
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