432. All O`one Data Structure (类似 LFU算法的设计)

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题目地址

https://leetcode.com/problems/all-oone-data-structure/?tab=Description

ac转载

http://www.cnblogs.com/grandyang/p/6012229.html

主要是双向链表 + map结构

数据结构设计

这里写图片描述

下面是纯粹用自己写的双向链表和 C++ map 的ac代码
需要注意各种inc和dec处理的各种情况

typedef struct node{    int val;    set<string> keys;    struct node* prev;    struct node* next;}Dnode;class AllOne {public:    Dnode* head;// 双向链表,头部存最大的，尾部存最小的    Dnode* tail;    map<string,Dnode*> m; // mappublic:    AllOne() {        head = new Dnode;        tail = new Dnode;        head->next = tail;        head->prev = NULL;        tail->prev = head;        tail->next = NULL;    }    void inc(string key) {        if(m.count(key) == 0) // 不存在        {            // 新插入的val就等于1，需要判断双向链表是否有val=1的             if(tail->prev == head || tail->prev->val != 1)            {                Dnode* dn = new Dnode;                dn->val = 1;                dn->keys.insert(key);                Dnode* tailpre = tail->prev;                tail->prev = dn;                dn->next = tail;                tailpre->next = dn;                dn->prev = tailpre;                m[key] = dn;            }else{                tail->prev->keys.insert(key);                m[key] = tail->prev;            }        }else{ // 存在，key所处的val需要加上1            Dnode* tmp = m[key];            Dnode* lastNode = tmp->prev;            if(lastNode == head || lastNode->val != tmp->val + 1)            {                Dnode* newDn = new Dnode;                newDn->val = tmp->val + 1;                newDn->keys.insert(key);                lastNode->next->prev = newDn;                newDn->next = lastNode->next;                newDn->prev = lastNode;                lastNode->next = newDn;                m[key] = newDn;            }else{                lastNode->keys.insert(key);                m[key] = lastNode;            }            tmp->keys.erase(key);            if(tmp->keys.empty())            {                Dnode* tmpPre = tmp->prev;                tmp->next->prev = tmpPre;                tmpPre->next = tmp->next;                delete tmp;            }        }    }    void dec(string key) {        if(m.count(key) == 0)            return;        Dnode* curDn = m[key];        if(curDn->val == 1)        {            curDn->keys.erase(key);            if(curDn->keys.empty())            {                Dnode* tmpPre = curDn->prev;                curDn->next->prev = tmpPre;                tmpPre->next = curDn->next;                delete curDn;                m.erase(key);                return;            }        }        Dnode* nextDn = curDn->next;        if(nextDn == tail || nextDn->val != curDn->val -1)        {            if(curDn->val - 1 == 0)            {                m.erase(key);            }else{                Dnode* newDn = new Dnode;                newDn->val = curDn->val - 1;                newDn->keys.insert(key);                curDn->next->prev = newDn;                newDn->next = curDn->next;                newDn->prev = curDn;                curDn->next = newDn;                m[key] = newDn;            }        }else{            nextDn->keys.insert(key);            m[key] = nextDn;        }        curDn->keys.erase(key);        if(curDn->keys.empty())        {            Dnode* tmpPre = curDn->prev;            curDn->next->prev = tmpPre;            tmpPre->next = curDn->next;            delete curDn;        }    }    string getMaxKey() {        if(head->next == tail)            return "";        return *(head->next->keys.begin());    }    string getMinKey() {        if(head->next == tail)            return "";        return *(tail->prev->keys.begin());    }};

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