[数据结构与算法]leetcode刷题——链表

1.leetcode203移除链表元素（简单）

题目连接：https://leetcode-cn.com/problems/remove-linked-list-elements/submissions/

解法：设置虚拟头节点

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode* virtualhead = new ListNode();
        virtualhead->next = head;
        ListNode* p = virtualhead;
        while (p->next != NULL)
        {
            if (p->next->val == val)
            {
                ListNode* tmp =  p->next;
                p->next = p->next->next;
                delete tmp;
            }
            else
                p = p->next;
        }
        head = virtualhead->next;
        delete virtualhead;
        return head;
    }
};

注意：记得释放需要删除的节点的内存。

2.leetcode707设计链表（中等）

题目链接：https://leetcode-cn.com/problems/design-linked-list/

#include<iostream>
using namespace std;

class MyLinkedList {
public:
    struct ListNode
    {
        int val;
        ListNode* next;
        ListNode(int val):val(val),next(NULL){}
    };
    ListNode* dummyHead;
    MyLinkedList() {
        dummyHead = new ListNode(0);
    }
    //获取链表中第index个节点的值
    int get(int index) {
        ListNode* p = dummyHead->next;
        int i = 0;
        while (p!= NULL && i < index)
        {
            p = p->next;
            i++;
        }
        if (p!= NULL) return p->val;
        else  return -1;
    }
    //在链表的第一个元素之前添加一个值为 val 的节点。插入后，新节点将成为链表的第一个节点。
    void addAtHead(int val) {
        ListNode* p = new ListNode(val);
        ListNode* ret = dummyHead->next;
        dummyHead->next = p;
        p->next = ret;
    }
    // addAtTail(val)：将值为 val 的节点追加到链表的最后一个元素。
    void addAtTail(int val) {
        ListNode* p = dummyHead;
        while (p->next != NULL)
        {
            p = p->next;
        }
        ListNode* newNode = new ListNode(val);
        p->next = newNode;
        newNode->next = NULL;
    }

    //在链表中的第 index 个节点之前添加值为 val  的节点。如果 index 等于链表的长度，则该节点将附加到链表的末尾。如果 index 大于链表长度，则不会插入节点。如果index小于0，则在头部插入节点。
    void addAtIndex(int index, int val) {
        if (index <= 0)
        {
            addAtHead(val);
            return;
        }
        ListNode* p = dummyHead;
        int i = 0;
        while (p!= NULL && i < index)
        {
            p = p->next;
            i++;
        }
        if (p!= NULL)
        {
            ListNode *newnode=new ListNode(val);
            newnode->next = p->next;
            p->next = newnode;
        }
    }
    
    //eleteAtIndex(index)：如果索引 index 有效，则删除链表中的第 index 个节点。
    void deleteAtIndex(int index) {
        if(index == 0) 
        {
            ListNode* p = dummyHead->next;
            dummyHead->next = p->next;
            delete p;
            return;
        }
        else if (index > 0)
        {
            ListNode* p = dummyHead;
            int i = 0;
            while (p!= NULL && i < index)
            {
                p = p->next;
                i++;
            }
            if(p == NULL)
            {
                return;
            }
            else if(p->next != NULL)
            {
            ListNode* ret = p->next;
            p->next = ret->next;
            delete ret;
            return;
            }
        }
    }
};

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * MyLinkedList* obj = new MyLinkedList();
 * int param_1 = obj->get(index);
 * obj->addAtHead(val);
 * obj->addAtTail(val);
 * obj->addAtIndex(index,val);
 * obj->deleteAtIndex(index);
 */

3.leetcode206反转链表（简单）

题目链接：https://leetcode-cn.com/problems/reverse-linked-list/

#include<iostream>
using namespace std;

//  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 == NULL || head->next == NULL) return head;
        else
        {
            ListNode* newhead = NULL;
            while (head != NULL)
            {
                ListNode* tmp = head->next;
                head->next = newhead;
                newhead = head;
                head = tmp;
            }
            return newhead;
        }

    }
};

4.leetcode24两两交换链表中的节点（中等）

题目链接：https://leetcode-cn.com/problems/swap-nodes-in-pairs/submissions/

解法一、递归法

 
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        if (head == NULL || head->next == NULL)
        {
            return head;
        }
        else
        {
            ListNode* newhead = head->next;
            head->next = swapPairs(newhead->next);
            newhead->next = head;
            return newhead;
        }
    }
};

解法二 非递归法