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AllOne.cpp
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277 lines (241 loc) · 8.31 KB
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class Node {
public:
string key;
int value;
Node* prev;
Node* next;
Node* up;
Node* down;
Node() : key(""), value(0), prev(NULL), next(NULL), up(NULL), down(NULL) {}
Node(string& _key) : key(_key), value(1), prev(NULL), next(NULL), up(NULL), down(NULL) {}
};
class HList {
public:
Node pseudoHead;
Node pseudoTail;
HList() {
pseudoHead.next = &pseudoTail;
pseudoTail.prev = &pseudoHead;
pseudoTail.value = INT_MAX;
}
void insert(Node* prevNode, Node* node, Node* nextNode) {
if (node->value == prevNode->value) {
prevNode->prev->next = node;
node->prev = prevNode->prev;
node->next = nextNode;
nextNode->prev = node;
node->down = prevNode;
prevNode->up = node;
prevNode->prev = prevNode->next = NULL;
}
else if (node->value == nextNode->value) { // 相同的要合并,放到当前节点下面
prevNode->next = node;
node->prev = prevNode;
node->next = nextNode->next;
nextNode->next->prev = node;
node->down = nextNode;
nextNode->up = node;
nextNode->prev = nextNode->next = NULL; // 纯纵向节点没有前后
}
else {
prevNode->next = node;
node->prev = prevNode;
node->next = nextNode;
nextNode->prev = node;
}
}
void remove(Node* node) {
Node* prevNode = node->prev;
Node* nextNode = node->next;
if (node->down == NULL) { // 这一列只剩这一个节点了,直接删除
prevNode->next = nextNode;
nextNode->prev = prevNode;
}
else { // 这一列还有其它节点,删了首节点第二个顶上来
prevNode->next = node->down;
node->down->prev = prevNode;
node->down->next = nextNode;
nextNode->prev = node->down;
}
}
};
class VList {
public:
Node pseudoTop;
Node pseudoBottom;
int size;
VList() {
pseudoTop.down = &pseudoBottom;
pseudoBottom.up = &pseudoTop;
size = 0;
}
void insert(Node* node) { // 只能插入在头部
node->up = &pseudoTop;
node->down = pseudoTop.down;
pseudoTop.down->up = node;
pseudoTop.down = node;
++size;
}
void remove(Node* node) {
node->up->down = node->down;
node->down->up = node->up;
if (--size == 0) {
node->up = node->down = NULL; // 删光之后要置空,因为删除横向的时候会判断纵向还有没有节点
}
}
};
class AllOne {
public:
/** Initialize your data structure here. */
AllOne() {
}
/** Inserts a new key <Key> with value 1. Or increments an existing key by 1. */
void inc(string key) {
if (keyNodes.count(key) == 0) {
Node* node = new Node(key);
keyNodes[key] = node;
insert(&hlist.pseudoHead, node, hlist.pseudoHead.next); // 新的节点肯定插入在pseudoHead后面
}
else {
Node* node = keyNodes[key];
Node* prev = valueLists[node->value].pseudoTop.down->prev;
Node* next = valueLists[node->value].pseudoTop.down->next;
remove(node); // 先删除这个节点
++node->value;
if (node->down == NULL) { // 原来那一列已经没有了
insert(prev, node, next);
}
else {
insert(valueLists[node->value - 1].pseudoTop.down, node, next);
}
}
}
/** Decrements an existing key by 1. If Key's value is 1, remove it from the data structure. */
void dec(string key) {
if (keyNodes.count(key) == 0) {
return;
}
Node* node = keyNodes[key];
Node* prev = valueLists[node->value].pseudoTop.down->prev;
Node* next = valueLists[node->value].pseudoTop.down->next;
remove(node); // 先删除这个节点
if (--node->value == 0) { // 减为0了就彻底干掉
keyNodes.erase(key);
delete node;
}
else { // 计数减1后再重新插入
if (node->down == NULL) { // 原来那一列已经没有了
insert(prev, node, next);
}
else {
insert(prev, node, valueLists[node->value + 1].pseudoTop.down);
}
}
}
/** Returns one of the keys with maximal value. */
string getMaxKey() {
return hlist.pseudoHead.next == &hlist.pseudoTail ? "" : hlist.pseudoTail.prev->key;
}
/** Returns one of the keys with Minimal value. */
string getMinKey() {
return hlist.pseudoHead.next == &hlist.pseudoTail ? "" : hlist.pseudoHead.next->key;
}
void remove(Node* node) {
VList& vlist = valueLists[node->value];
vlist.remove(node); // 先从纵向队列删除
if (vlist.size == 0) {
valueLists.erase(node->value);
}
if (node->prev != NULL) {
hlist.remove(node); // 再从横向队列删除
}
}
void insert(Node* prev, Node* node, Node* next) {
valueLists[node->value].insert(node);
hlist.insert(prev, node, next);
}
private:
unordered_map<string, Node*> keyNodes;
unordered_map<int, VList> valueLists;
HList hlist;
};
// 作者:wen-rou-yi-dao-123
// 链接:https://leetcode-cn.com/problems/all-oone-data-structure/solution/shi-zi-lian-biao-wan-cheng-quan-o1cao-zu-iww6/
// 来源:力扣(LeetCode)
// 著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
/**
* Your AllOne object will be instantiated and called as such:
* AllOne* obj = new AllOne();
* obj->inc(key);
* obj->dec(key);
* string param_3 = obj->getMaxKey();
* string param_4 = obj->getMinKey();
*/
class AllOne {
public:
/** Initialize your data structure here. */
struct Node{
unordered_set<string> container;
int val = 0;
Node(int v):val(v){}
};
unordered_map<string, list<Node>::iterator> kv;
list<Node> List;
AllOne() {}
/** Inserts a new key <Key> with value 1. Or increments an existing key by 1. */
void inc(string key) {
if(kv.count(key)){
auto oldNode = kv[key];
auto newNode = next(oldNode, 1);
if(newNode == List.end() || newNode->val>oldNode->val+1){
newNode = List.insert(newNode, Node(oldNode->val+1));
}
newNode->container.insert(key);
oldNode->container.erase(key);
if(oldNode->container.empty()){
List.erase(oldNode);
}
kv[key] = newNode;
} else {
auto newNode = List.begin();
if(List.empty() || List.begin()->val>1)
newNode = List.insert(List.begin(), Node(1));
newNode->container.insert(key);
kv[key] = newNode;
}
}
/** Decrements an existing key by 1. If Key's value is 1, remove it from the data structure. */
void dec(string key) {
if(kv.count(key)){
auto oldNode = kv[key];
if(oldNode->val==1) {
kv.erase(key);
} else {
auto newNode = next(oldNode, -1);
if(oldNode==List.begin() || newNode->val<oldNode->val-1){
newNode = List.insert(oldNode, Node(oldNode->val-1));
}
newNode->container.insert(key);
kv[key] = newNode;
}
oldNode->container.erase(key);
if(oldNode->container.empty()){
List.erase(oldNode);
}
}
}
/** Returns one of the keys with maximal value. */
string getMaxKey() {
if(List.empty()) return "";
return *List.rbegin()->container.begin();
}
/** Returns one of the keys with Minimal value. */
string getMinKey() {
if(List.empty()) return "";
return *List.begin()->container.begin();
}
};
// 作者:lwpyr
// 链接:https://leetcode-cn.com/problems/all-oone-data-structure/solution/c-ha-xi-kvshuang-xiang-lian-biao-by-lwpyr/
// 来源:力扣(LeetCode)
// 著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。