左神基础班-单链表存储的数，判断回文
先上时间复杂度和空间复杂度 都是 O(n)的解法
里面的几个函数功能分别为：1.由数组生成链表、2.得到反转后的链表、3.顺序打印链表，4.比较两个链表，若有一处不同，则不是回文。
#include
#include
using namespace std;
struct LinkNode{
int value;
LinkNode* next;
};
LinkNode* getLink(int arr[], int length);
LinkNode* getLink_reverse(LinkNode* head1, int length);
void show(LinkNode* head);
bool compare(LinkNode* head1, LinkNode* head2);
int main(){
int arr1[] = {1,2,3,2,1};
LinkNode* head1 = getLink(arr1,5);
LinkNode* head2 = getLink_reverse(head1, 5);
// show(head1);
// show(head2);
cout << "is huiwen: "<< compare(head1, head2);
return 0;
}
LinkNode* getLink(int arr[], int length){
LinkNode* head = new LinkNode();
LinkNode* go = head;
int i =0;
while(ivalue = arr[i];
node->next = nullptr;
go->next = node;
go = go->next;
i++;
}
return head;
}
LinkNode* getLink_reverse(LinkNode* head1 , int length){
stack s;
LinkNode* go = head1->next;
while(go!=nullptr){
s.push(go->value);
go=go->next;
}
LinkNode* head2 = new LinkNode();
LinkNode* go2 = head2;
while(!s.empty()){
LinkNode* node = new LinkNode();
node->value = s.top();
s.pop();
node->next = nullptr;
go2->next = node;
go2 = go2->next;
}
return head2;
}
void show(LinkNode* head){
LinkNode* go = head->next;
while(go!=nullptr){
cout << go->value << endl;
go = go->next;
}
}
bool compare(LinkNode* head1, LinkNode* head2){
LinkNode* go1 = head1->next;
LinkNode* go2 = head2->next;
while((go1!= nullptr && go2 != nullptr) &&
go1->value == go2->value){
go1 = go1->next;
go2 = go2->next;
}
return go1==nullptr? true : false;
} +————————————————————————————————————————————————————————+
下面贴时间复杂度位O(n)，空间复杂度为O(1)的代码
#include
#include
using namespace std;
struct LinkNode{
int value;
LinkNode* next;
};
LinkNode* getLink(int arr[], int length);
void show(LinkNode* head);
bool judge(LinkNode* head);
int main(){
int arr1[] = {1,2,3,3,2,1};
LinkNode* head1 = getLink(arr1,6);
// show(head1);
cout << "is huiwen: " << judge(head1)<value = arr[i];
node->next = nullptr;
go->next = node;
go = go->next;
i++;
}
return head;
}
void show(LinkNode* head){
LinkNode* go = head->next;
while(go!=nullptr){
cout << go->value << endl;
go = go->next;
}
}
bool judge(LinkNode* head){
LinkNode* slow;
LinkNode* fast;
slow = fast = head->next;
while(fast->next != nullptr && fast->next->next != nullptr){
slow = slow->next;
fast = fast->next->next;
}
//此时，slow指向中间的节点
fast = slow->next;
slow->next = nullptr;
LinkNode* help = fast->next;
while(help != nullptr){
fast->next = slow;
slow = fast;
fast = help;
help = help->next;
}
fast->next = slow;
//反转指针完毕,判断回文情况
LinkNode* front = head->next;
LinkNode* tail = fast;
while(front !=nullptr ){
if(front->value != tail->value){
return false;
}
// cout << "now ,front is:" << front->value <<" tail is: "<< tail->value << endl;
front = front->next;
tail = tail->next;
}
//把指针调整回去
slow = fast;
fast = fast->next;
slow->next = nullptr;
help = fast->next;
while(help != nullptr){
fast->next = slow;
slow = fast;
fast = help;
help = help->next;
}
fast->next = slow;
return true;
}
来个图片帮助理解。
其中，要设立一个help指针。其指向正在被改变指针节点 的那个节点的下一个节点。
这样，当一个节点其后指针改变时。还能找到其原先的后一个节点，方能继续修改。