【算法训练营day3】LeetCode203. 移除链表元素 707. 设计链表 206. 反转链表

LeetCode203. 移除链表元素#

题目链接：203. 移除链表元素

初次尝试#

题目比较简单，之前刷过链表的题，有一段时间不碰了，刚开始对语法有一点生疏，后面找到感觉了，一遍ac。

/**
 * 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* newHead = new ListNode(0, head);
        ListNode* p = newHead;
        ListNode* temp;

        while (p -> next != NULL) {
            if (p -> next -> val == val) {
                temp = p -> next;
                p -> next = temp -> next;
                delete temp;
            }
            else p = p -> next;
        }
        
        return newHead -> next;
    }
};

看完代码随想录后的想法#

思路是一样的，感觉虚拟头节点的代码会更简单，思考量也会更小。

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LeetCode707. 设计链表#

题目链接：707. 设计链表

初次尝试#

思考量不大的一道题，比较考验基本功，一遍ac。

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

    MyLinkedList() {
        _newHead = new LinkedNode(0);
        _size = 0;
    }
    
    int get(int index) {
        LinkedNode* p = _newHead;

        for (; index >= 0; index--) {
            p = p -> next;
            if (p == NULL) return -1;
        }

        return p -> val;
    }
    
    void addAtHead(int val) {
        LinkedNode* newNode = new LinkedNode(val);
        newNode -> next = _newHead -> next;
        _newHead -> next = newNode;
        _size++;
    }
    
    void addAtTail(int val) {
        LinkedNode* newNode = new LinkedNode(val);
        LinkedNode* p = _newHead;

        for (int i = 0; i < _size; i++) {
            p = p -> next;
        }

        p -> next = newNode;
        _size++;
    }
    
    void addAtIndex(int index, int val) {
        if (index < 0) addAtHead(val);
        LinkedNode* newNode = new LinkedNode(val);
        LinkedNode* p = _newHead;

        for (; index > 0; index--) {
            p = p -> next;
            if (p == NULL) return;
        }

        newNode -> next = p -> next;
        p -> next = newNode;
        _size++;
    }
    
    void deleteAtIndex(int index) {
        LinkedNode* p = _newHead;
        LinkedNode* temp;

        for (; index > 0; index--) {
            p = p -> next;
            if (p -> next == NULL) return;
        }

        temp = p -> next;
        p -> next = temp -> next;
        delete temp;
        _size--;
    }

private:
    LinkedNode* _newHead;
    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);
 */

看完代码随想录后的想法#

思路一样。

---

LeetCode206. 反转链表#

题目链接：206. 反转链表

初次尝试#

我的想法是建一个新的链表，然后遍历旧链表，依次插入到新链表中，一遍ac。

/**
 * 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) {
        if (head == NULL) return NULL;
        ListNode* ans = new ListNode();
        ListNode* p = head;
        ListNode* temp = p -> next;

        while (p != NULL) {
            p -> next = ans -> next;
            ans -> next = p;
            p = temp;
            if (temp == NULL) break;
            temp = temp -> next;
        }

        return ans -> next;
    }
};

看完代码随想录后的想法#

题解的方法非常巧妙，确实只需要两个指针遍历链表将next指针反向就行了。

/**
 * 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* cur = head;
        ListNode* pre = NULL;
        ListNode* temp;

        while (cur) {
            temp = cur -> next;
            cur -> next = pre;
            pre = cur;
            cur = temp;
        }
        
        return pre;
    }
};