代码随想录算法训练营day03|203.移除链表元素，707.设计链表，206.反转链表

合集 - 代码随想录算法训练营C++(8)

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3.代码随想录算法训练营day03|203.移除链表元素，707.设计链表，206.反转链表2024-08-03

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203.移除链表元素

题目链接：https://leetcode.cn/problems/remove-linked-list-elements/description/

我的代码（分头节点和中间节点两种情况操作）：

/**
 * 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* removeElements(ListNode* head, int val) {
        while (head != nullptr && head->val == val) {
            ListNode* q = head;
            head = head->next;
            delete q;
        }
        ListNode* p = head;
        while (p != nullptr && p->next != nullptr) {
            if (p->next->val == val) {
                ListNode* q = p->next;
                p->next = q->next;
                delete q;
            } else
                p = p->next;
        }
        return head;
    }
};

头节点操作：直接后移头节点并将原来的头节点内存释放。
中间节点操作：检测到当前节点后一个节点的数据域等于索引值时，将当前节点的指针域指向后一个节点的后一个节点，然后释放后一个节点的内存，否则后移当前节点。

虚拟头节点（所有节点按一种方法操作）：

/**
 * 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* removeElements(ListNode* head, int val) {
        ListNode* dummy_head = new ListNode;
        dummy_head->next = head;
        ListNode* current = dummy_head;
        while (current->next != nullptr) {
            if (current->next->val == val) {
                ListNode* temp = current->next;
                current->next = current->next->next;
                delete temp;
            } else
                current = current->next;
        }
        head = dummy_head->next;
        delete dummy_head;
        return head;
    }
};

在原链表的头节点之前再加一个实际上不存储数据的虚拟头节点，方便操作。

707.设计链表

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

我的代码：

class MyLinkedList {
public:
    struct LinkedList {
        int val;
        struct LinkedList* next;
        LinkedList(int val) : val(val), next(nullptr) {}
    };
    MyLinkedList() {
        dummy_head = new LinkedList(0);
        size = 0;
    }
 
    int get(int index) {
        if (index < size && index >= 0) {
            LinkedList* p = dummy_head->next;
            while (index > 0) {
                index--;
                p = p->next;
            }
            return p->val;
        } else
            return -1;
    }
 
    void addAtHead(int val) {
        LinkedList* p = new LinkedList(val);
        p->next = dummy_head->next;
        dummy_head->next = p;
        size++;
    }
 
    void addAtTail(int val) {
        LinkedList* p = new LinkedList(val);
        LinkedList* q = dummy_head;
        int length = size;
        while (length > 0) {
            length--;
            q = q->next;
        }
        q->next = p;
        p->next = nullptr;
        size++;
    }
 
    void addAtIndex(int index, int val) {
        if (index >= 0 && index < size) {
            LinkedList* p = new LinkedList(val);
            LinkedList* q = dummy_head;
            while (index > 0) {
                index--;
                q = q->next;
            }
            p->next = q->next;
            q->next = p;
            size++;
        } else if (index == size) {
            LinkedList* p = new LinkedList(val);
            LinkedList* q = dummy_head;
            while (index > 0) {
                index--;
                q = q->next;
            }
            q->next = p;
            p->next = nullptr;
            size++;
        }
    }
 
    void deleteAtIndex(int index) {
        LinkedList* p = dummy_head;
        if (index < size && index >= 0) {
            while (index > 0) {
                index--;
                p = p->next;
            }
            LinkedList* q = p->next;
            p->next = q->next;
            delete q;
            size--;
        }
    }
 
private:
    LinkedList* dummy_head;
    int size;
};
 
/**
 * 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);
 */

206.反转链表

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

双指针解法：

/**
 * 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* reverseList(ListNode* head) {
        ListNode* pre = nullptr;
        ListNode* current = head;
        while (current != nullptr) {
            ListNode* temp = current->next;
            current->next = pre;
            pre = current;
            current = temp;
        }
        return pre;
    }
};

pre：开始定义为头节点前的空指针。
current：开始定义为头节点。
每次循环使用temp暂存current的next节点，然后使current的next指针指向pre，pre和current都后移，最后当current指向尾节点后的空指针时结束循环，返回当前真正头节点pre。

递归解法：

/**
 * 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* reverseList(ListNode* head) { return reverse(nullptr, head); }
    ListNode* reverse(ListNode* pre, ListNode* current) {
        if (current == nullptr)
            return pre;
        ListNode* temp = current->next;
        current->next = pre;
        return reverse(current, temp);
    }
};

与双指针解法思路相同。