DAY3|LeetCode203.移除链表元素 、707.设计链表、206.反转链表

203.移除链表元素

这一道是链表比较基础的一题，为了统一特殊情况规则（删除头节点的规则和中间节点的不一），可以设置一个虚拟头结点，代码如下所示：

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode* dummyHead = new ListNode(0);//创虚拟头结点
        dummyHead->next = head; 
        ListNode* cur = dummyHead;
        while(cur->next != NULL)
        {
            if(cur->next->val ==val )
            {
                ListNode* tmp = cur->next;
                cur->next = cur->next->next;
                delete tmp;
            }
            else 
                cur = cur->next;
        }
    head = dummyHead->next;
    delete dummyHead;
    return head;  

    }
};

707.设计链表

class MyLinkedList {
public:
    struct LinkedNode {
        int val;
        LinkedNode* next;
        LinkedNode(int val):val(val), next(nullptr){}
    };

    MyLinkedList() {
        _dummyHead = new LinkedNode(0); 
        _size = 0;
    }
    int get(int index) {
        if (index > (_size - 1) || index < 0) {
            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;
        _size--;
    }

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

};

206.反转链表

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* temp; 
        ListNode* cur = head;
        ListNode* pre = NULL;
        while(cur!=NULL) {
            temp = cur->next;  
            cur->next = pre; 
            pre = cur;
            cur = temp;
        }
        return pre;
    }
};

这道题采用的是双指针法，首先定义一个cur指针，指向头结点，再定义一个pre指针，初始化为NULL，当cur不等于NULL时，改变cur->next的指向，最后返回pre。