LeetCode——707 设计链表

题目：

总而言之就是要用C++手撸链表，我的代码：

class MyLinkedList {
public:
    /** Initialize your data structure here. */
    MyLinkedList() {
        head = NULL;
        size = 0;
    }

    /** Get the value of the index-th node in the linked list. If the index is invalid, return -1. */
    int get(int index) {
        if(index<0 || index >= size){
            return -1;
        }
        Node *cur = head;
        for(int i=0; i<index; i++){
            cur = cur->next;
        }
        return cur->val;
    }

    /** Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list. */
    void addAtHead(int val) {
        Node *newhead = new Node(val, head);
        head = newhead;
        size++;
    }

    /** Append a node of value val to the last element of the linked list. */
    void addAtTail(int val) {
        Node *tail = new Node(val, NULL);
        Node *cur = head;
        while(cur->next)cur = cur->next;
        cur->next = tail;
        size++;
    }

    /** Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted. */
    void addAtIndex(int index, int val) {
        if(index>size){
            return;
        }
        if(index==0||index==-1){
            addAtHead(val);
            return;
        }
        if(index==size){
            addAtTail(val);
            return;
        }
        Node *cur = head;
        for(int i=0; i<index-1; i++){
            cur = cur->next;
        }
        Node *toAdd = new Node(val, cur->next);
        cur->next = toAdd;
        size++;
    }

    /** Delete the index-th node in the linked list, if the index is valid. */
    void deleteAtIndex(int index) {
        if(index<0 || index >= size) return;
        Node *cur = head;
        if(index == 0){
            head = head->next;
            size--;
            delete cur;
            return;
        }
        for(int i=0; i<index-1; i++){
            cur = cur->next;
        }
        Node *toFree = cur->next;
        cur->next = cur->next->next;
        delete toFree;
        size--;
    }

private:
    struct Node{
        int val;
        Node *next;
        Node(int x, Node *n): val(x), next(n) {};
    };
    Node *head;
    int size;
};

这题其实很简单，这里的实现方式是单链表，基本上就是数据结构的知识点，所以没什么好提的。但是万恶的LeetCode有这么个测试样例：

也就是说，当输入的index为-1时要当成0来处理，但是这一点题目里面完全没有提及。只需要改一下判定条件就可以了（把void addAtIndex(int index, int val)里的if(index==0)改成if(index==0||index==-1)），除了这个问题之外基本上没什么要注意的了。