代码随想录笔记--链表篇

目录

1--虚拟头节点的使用

2--设计链表

3--反转链表

4--两两交换链表中的节点

5--快慢指针

5-1--删除链表倒数第N个节点

5-2--环形链表

5-3--环形链表II

---

1--虚拟头节点的使用

        在链表相关题目中，常新定义一个虚拟头结点 dummynode 来指向原链表的头结点，当需要返回链表时，只需返回 dummynode->next;

#include <iostream>

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* removeElements(ListNode* head, int val) {
        if(head==nullptr) return head;
        ListNode* dummynode = new ListNode(); // 新建虚拟头结点
        dummynode->next = head;
        ListNode* cur = dummynode;
        while(cur->next != nullptr){
            if(cur->next->val == val){
                ListNode* tmp = cur->next;
                cur->next = cur->next->next;
                delete tmp;
                tmp = nullptr;
            }
            else{
                cur = cur->next;
            }
        }
        return dummynode->next;
    }
};

int main(int arc, char *argv[]){
    ListNode* Node1 = new ListNode(1);
    ListNode* Node2 = new ListNode(2);
    ListNode* Node3 = new ListNode(6);
    ListNode* Node4 = new ListNode(3);
    ListNode* Node5 = new ListNode(4);
    ListNode* Node6 = new ListNode(5);
    ListNode* Node7 = new ListNode(6);

    Node1->next = Node2;
    Node2->next = Node3;
    Node3->next = Node4;
    Node4->next = Node5;
    Node5->next = Node6;
    Node6->next = Node7;

    Solution S1;
    int val = 6;
    ListNode* res = S1.removeElements(Node1, val);
    ListNode* cur = res;
    while(cur != NULL){
        std::cout << cur->val << " ";
        cur = cur->next;
    }
    return 0;
}

2--设计链表

#include <iostream>

class MyLinkedList {
public:
    // 定义链表节点结构体
    struct LinkedNode {
        int val;
        LinkedNode* next;
        LinkedNode(int val):val(val), next(nullptr){}
    };

    // 初始化链表
    MyLinkedList() {
        _dummyHead = new LinkedNode(0); // 虚拟头结点
        _size = 0;
    }

    int get(int index) {
        if (index > (_size - 1) || index < 0) { // 不合法index
            return -1;
        }
        LinkedNode* cur = _dummyHead->next;
        while(index--){
            cur = cur->next;
        }
        return cur->val;
    }

    void addAtHead(int val) {
        LinkedNode* newNode = new LinkedNode(val);
        newNode->next = _dummyHead->next;
        _dummyHead->next = newNode;
        _size++;
    }

    void addAtTail(int val) {
        LinkedNode* newNode = new LinkedNode(val);
        LinkedNode* cur = _dummyHead;
        while(cur->next != nullptr){
            cur = cur->next;
        }
        cur->next = newNode;
        _size++;
    }

    void addAtIndex(int index, int val) {

        if(index > _size) return;
        if(index < 0) index = 0;        
        LinkedNode* newNode = new LinkedNode(val);
        LinkedNode* cur = _dummyHead;
        while(index--) {
            cur = cur->next;
        }
        newNode->next = cur->next;
        cur->next = newNode;
        _size++;
    }

    void deleteAtIndex(int index) {
        if (index >= _size || index < 0) {
            return;
        }
        LinkedNode* cur = _dummyHead;
        while(index--) {
            cur = cur ->next;
        }
        LinkedNode* tmp = cur->next;
        cur->next = cur->next->next;
        delete tmp;
        tmp=nullptr;
        _size--;
    }

    // 打印链表
    void printLinkedList() {
        LinkedNode* cur = _dummyHead;
        while (cur->next != nullptr) {
            std::cout << cur->next->val << " ";
            cur = cur->next;
        }
        std::cout << std::endl;
    }
private:
    int _size;
    LinkedNode* _dummyHead;

};

int main(int arc, char *argv[]){
    MyLinkedList myLinkedList;
    myLinkedList.addAtHead(1);
    myLinkedList.addAtTail(3);
    myLinkedList.addAtIndex(1, 2);    // 链表变为 1->2->3
    std::cout << myLinkedList.get(1) << std::endl;  // 返回 2
    myLinkedList.deleteAtIndex(1);    // 现在，链表变为 1->3
    std::cout << myLinkedList.get(1) << std::endl;  // 返回 3
    return 0;
}

3--反转链表

双指针解法： 

#include <iostream>

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* reverseList(ListNode* head){
        if(head == nullptr) return head;
        ListNode *pre = nullptr;
        ListNode *cur = head;
        while(cur != NULL){
            ListNode *tmp = cur->next;
            cur->next = pre;
            pre = cur;
            cur = tmp;
        }
        return pre;
    }
};

int main(int arc, char *argv[]){
    ListNode *Node1 = new ListNode(1);
    ListNode *Node2 = new ListNode(2);
    ListNode *Node3 = new ListNode(3);
    ListNode *Node4 = new ListNode(4);
    ListNode *Node5 = new ListNode(5);

    Node1->next = Node2;
    Node2->next = Node3;
    Node3->next = Node4;
    Node4->next = Node5;

    Solution S1;
    ListNode *res = S1.reverseList(Node1);
    ListNode *cur = res;
    while(cur != NULL){
        std::cout << cur->val << " ";
        cur = cur->next;
    }
    return 0;
}

递归写法：

#include <iostream>

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* reverseList(ListNode* head){
        if(head == nullptr) return head;
        return reverse(head, nullptr);
    }

    ListNode* reverse(ListNode* cur, ListNode* pre){
        if(cur == nullptr) return pre;
        ListNode *tmp = cur->next;
        cur->next = pre;
        return reverse(tmp, cur);
    }
};

int main(int arc, char *argv[]){
    ListNode *Node1 = new ListNode(1);
    ListNode *Node2 = new ListNode(2);
    ListNode *Node3 = new ListNode(3);
    ListNode *Node4 = new ListNode(4);
    ListNode *Node5 = new ListNode(5);

    Node1->next = Node2;
    Node2->next = Node3;
    Node3->next = Node4;
    Node4->next = Node5;

    Solution S1;
    ListNode *res = S1.reverseList(Node1);
    ListNode *cur = res;
    while(cur != NULL){
        std::cout << cur->val << " ";
        cur = cur->next;
    }
    return 0;
}

4--两两交换链表中的节点

#include <iostream>

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* swapPairs(ListNode* head) {
        if(head == nullptr) return head;
        ListNode *dummyNode = new ListNode();
        dummyNode->next = head;
        ListNode *cur = dummyNode;
        while(cur->next != nullptr && cur->next->next != nullptr){
            ListNode *tmp1 = cur->next;
            ListNode *tmp2 = cur->next->next->next;
            cur->next = cur->next->next;
            cur->next->next = tmp1;
            tmp1->next = tmp2;
            cur = cur->next->next;
        }
        return dummyNode->next;
    }
};

int main(int arc, char *argv[]){
    ListNode *Node1 = new ListNode(1);
    ListNode *Node2 = new ListNode(2);
    ListNode *Node3 = new ListNode(3);
    ListNode *Node4 = new ListNode(4);

    Node1->next = Node2;
    Node2->next = Node3;
    Node3->next = Node4;

    Solution S1;
    ListNode *res = S1.swapPairs(Node1);
    ListNode *cur = res;
    while(cur != NULL){
        std::cout << cur->val << " ";
        cur = cur->next;
    }
    return 0;
}

5--快慢指针

5-1--删除链表倒数第N个节点

快慢指针

#include <iostream>
#include <vector>

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) {
        ListNode *dummyNode = new ListNode();
        dummyNode->next = head;
        ListNode* slow = dummyNode;
        ListNode* fast = dummyNode;
        while(n-- && fast != NULL){
            fast = fast->next;
        }
        while(fast->next != NULL){
            slow = slow->next;
            fast = fast->next;
        }
        ListNode *target = slow->next;
        slow->next = target->next;
        delete target;
        target = nullptr;
        return dummyNode->next;
    }
};

int main(int argc, char argv[]){
    ListNode *Node1 = new ListNode(1);
    ListNode *Node2 = new ListNode(2);
    ListNode *Node3 = new ListNode(3);
    ListNode *Node4 = new ListNode(4);
    ListNode *Node5 = new ListNode(5);

    Node1->next = Node2;
    Node2->next = Node3;
    Node3->next = Node4;
    Node4->next = Node5;

    int n = 2;
    Solution S1;
    ListNode *res = S1.removeNthFromEnd(Node1, n);
    ListNode *cur = res;
    while(cur != NULL){
        std::cout << cur->val << " ";
        cur = cur->next;
    }
    return 0;
}

5-2--环形链表

快慢指针，当链表有环时，快慢指针必定相遇；

#include <iostream>
#include <vector>

struct ListNode {
     int val;
     ListNode *next;
     ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    bool hasCycle(ListNode *head) {
        if(head == nullptr) return false;
        ListNode *slow = head;
        ListNode *fast = head;
        while(fast != NULL && fast->next != NULL){
            fast = fast->next->next;
            slow = slow->next;
            if(slow == fast) break;
        }
        if(fast == NULL || fast->next == NULL) return false;
        else return true;
    }
};

int main(int argc, char argv[]){
    ListNode *Node1 = new ListNode(1);
    ListNode *Node2 = new ListNode(2);
    ListNode *Node3 = new ListNode(0);
    ListNode *Node4 = new ListNode(-4);

    Node1->next = Node2;
    Node2->next = Node3;
    Node3->next = Node4;
    Node4->next = Node2;

    Solution S1;
    bool res = S1.hasCycle(Node1);
    if(res) std::cout << "true" << std::endl;
    else std::cout << "false" << std::endl;
    return 0;
}

5-3--环形链表II

        快慢指针，快指针每次走两步，慢指针每次走一步，根据快慢指针是否相遇判断是否有环；

        当快慢指针第一次相遇后，慢指针从头出发，快指针从第一次相遇的节点出发，快慢指针均每次走一步，当下一次相遇时就处于环的入口节点，返回即可；

#include <iostream>
#include <vector>

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if(head == nullptr) return head;
        ListNode *slow = head;
        ListNode *fast = head;
        while(fast != NULL && fast->next != NULL){
            fast = fast->next->next;
            slow = slow->next;
            if(fast == slow) break; // 第一次相遇
        }
        if(fast == NULL || fast->next == NULL) return NULL; // 没有环
        // 从头开始走
        slow = head;
        while(slow != fast){
            slow = slow->next;
            fast = fast->next;
        }
        // 第二次相遇就是环的入口
        return slow;
    }
};

int main(int argc, char argv[]){
    ListNode *Node1 = new ListNode(1);
    ListNode *Node2 = new ListNode(2);
    ListNode *Node3 = new ListNode(0);
    ListNode *Node4 = new ListNode(-4);

    Node1->next = Node2;
    Node2->next = Node3;
    Node3->next = Node4;
    Node4->next = Node2;

    Solution S1;
    ListNode * res = S1.detectCycle(Node1);
    std::cout << res->val << std::endl;
    return 0;
}