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c++数据结构算法复习基础-- 3 --线性表-单向链表-笔试面试常见问题

news 2025/8/15 2:27:57

1、单链表逆序

思路图

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代码实现

//著: 链表结构里记得加 friend void ReverseLink(Clink& link);
void ReverseLink(Clink& link)
{Node* p = link.head_->next_;while( p == nullptr){return;}Node* q = p->next_;link.head_->next_ = nullptr;while(p != nullptr){Node* q = p->next_;//p指针指向的节点进行头插p->next_ = link.head_->next_;link.head_->next_ = p;p = q;}/*Node* p = link.head_->next_;if( p == nullptr){return;}Node* q = p->next_;link.head_->next_ = nullptr;while( q != nullptr){link.head_->data_ = p->data_;p->next_ = link.head_;link.head_ = p;p = q;q = p->next_;}link.head_->data_ = p->data_;p->next_ = link.head_;link.head_ = p;p = nullptr;*/
}

测试

int main()
{Clink link;Clink link2;srand(time(0));for(int i = 0; i<10; i++){int val = rand() % 100;link.InsertTail(val);}link.show();cout << endl;ReverseLink(link);ReverseLink(link2);link.show();cout << endl;link2.show();cout << endl;
}

运行结果

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2、单链表倒数第k个节点

思路图

双指针同步位移，两个指针相聚k个节点
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代码实现

//求倒数第k个节点的值
bool MyGetLastKNode(Clink& link,int k,int& val)
{if(k<1){return 0;}Node* p = link.head_;if( p == nullptr){return false;}Node* pre = link.head_;for(int i = 0; i < k ; i++){pre = pre->next_;if( pre == nullptr ){return false;}}//p在头节点，pre在正数第k个节点while( pre != nullptr ){p = p->next_;pre = pre->next_;}val = p->data_;return true;
}

代码测试

int main()
{Clink link;Clink link2;srand(time(0));for(int i = 0; i<10; i++){int val = rand() % 100;link.InsertTail(val);}link.show();cout << endl;int val=0;if(MyGetLastKNode(link,3,val)){cout<< "k == 3   ";cout<< "find  " << val<< endl;}else{cout<< "k == 3   ";cout<< "false" << endl;}if(MyGetLastKNode(link,0,val)){cout<< "k == 0   ";cout<< "find  " << val<< endl;}else{cout<< "k == 0   ";cout<< "false" << endl;}if(MyGetLastKNode(link,12,val)){cout<< "k == 12   ";cout<< "find  " << val<< endl;}else{cout<< "k == 12   ";cout<< "false" << endl;}if(MyGetLastKNode(link2,3,val)){cout<< "k2 == 3   ";cout<< "find  " << val<< endl;}else{cout<< "k2 == 3   ";cout<< "false" << endl;}}

运行结果

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3、并两个有序单链表

思路图

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代码实现

//合并两个有序单链表
bool MergeLink(Clink& link1,Clink& link2)
{Node* p = link1.head_->next_;Node* q = link2.head_->next_;Node* last = link1.head_;link2.head_->next_ = nullptr;while(p != nullptr && q != nullptr){if(p->data_ < q->data_){last->next_ = p;p = p->next_;last = last->next_;}else{last->next_ = q;q = q->next_;last = last->next_;}}if(p != nullptr){last->next_ = p;}else{last->next_ = q;}/*Node* pre = link1.head_;Node* p = link1.head_->next_;Node* q = link2.head_->next_;while(q != nullptr){if(p == nullptr && q != nullptr){pre->next_ = q;link2.head_->next_ = nullptr;return true;}else if(p->data_ <= q->data_)//这里假设从小到大{p = p->next_;pre = pre->next_;}else {link2.head_->next_ = q->next_;pre->next_=q;q->next_ = p;pre=q;q = link2.head_->next_;}}*/return true;
}

运行结果

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4、判断单链表是否存在环以及入口节点

思路图

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代码实现

//判断单链表是否存在环以及入口节点
//这里参数为Node,方便测试
//记得 friend
bool IsLinkHasCirle(Node* head,int& val)
{Node* fast = head;Node* slow = head;while(fast != nullptr && fast->next_ != nullptr){slow = slow->next_;fast = fast->next_->next_;if(slow == fast){//快慢指针再次相遇，链表存在环fast = head;while(fast != slow){slow = slow->next_;fast = fast->next_;}val = slow->data_;return true;}}return false;
}

测试

int main()
{Node head;Node n1(25),n2(61),n3(312),n4(118);head.next_ = &n1;n1.next_ = &n2;n2.next_ = &n3;n3.next_ = &n4;n4.next_ = &n2;int val;if(IsLinkHasCirle(&head,val)){cout<< "链表存在环，环的入口节点是： "<< val << endl;}}

运行结果

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5、判断两个链表是否相交

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思路图

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代码实现

//判断两个链表是否相交,如果相交，返回相交节点的值
bool IsLinkHasMerge(Node* head1,Node* head2,int &val)
{int cnt1 = 0,cnt2 = 0;Node* p = head1->next_;Node* q = head2->next_;//计算两个链表的长度while(p != nullptr){p = p->next_;cnt1++;}while(q != nullptr){q = q->next_;cnt2++;}p = head1->next_;q = head2->next_;if(cnt1 > cnt2){//第一个链表长cnt1 = cnt1-cnt2;while(cnt1 >0){p = p->next_;cnt1--;}}else{//第二个链表长cnt2 = cnt2 -cnt1;while(cnt2 > 0){q = q->next_;cnt2--;}}while(p != nullptr && q != nullptr){if(p == q){val = p->data_;return true;}p = p->next_;q = q->next_;}return false;
}

测试代码

int main()
{Node head1;Node head2;Node n1(25),n2(61),n3(312),n4(118),n5(18);Node nn1(1),nn2(2);head1.next_ = &n1;n1.next_ = &n2;n2.next_ = &n3;n3.next_ = &n4;n4.next_ = &n5;head2.next_ = &nn1;nn1.next_ = &nn2;nn2.next_ = &n3;int val;if(IsLinkHasMerge(&head1,&head2,val)){cout<< "两个链表相交，相交的节点是： "<< val << endl;}}

运行结果

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6、删除链表倒数第N个节点

思维图

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代码实现

//这里使用利扣刷题
/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* removeNthFromEnd(ListNode* head, int n) {//在函数内部给链表增加一个头节点，以解决不带头节点的单链表//head_->next   headListNode* head_ = new ListNode(0,head);ListNode* first = head;ListNode* second = head_;for(int i = 0; i < n; i++){first = first->next;}while(first){first = first->next;second = second->next;}second->next = second->next->next;ListNode* ans = head_->next;delete head_;return ans;;}};

7、旋转链表

思路图

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代码实现

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* rotateRight(ListNode* head, int k) {ListNode*p = head;ListNode*q = head;if(head == nullptr || k == 0){return head;}int number = 0;//判断链表的长度for(ListNode *k = head; k != nullptr; k = k->next){number++;}k = k%number;for(int i = 0; i < k; i++){p = p->next;}while(p->next != nullptr){q = q->next;p = p->next;}p->next = head;head = q->next;q->next = nullptr;return head;}
};