问题描述
- 跪求大神 c++词频 排序
-
You say that you love rain,
but you open your umbrella when it rains.
You say that you love the sun,
but you find a shadow spot when the sun shines.
You say that you love the wind,
but you close your windows when wind blows.
This is why I am afraid,
because you say that you love me too。
将上述文字 利用c++堆栈 指针 或者数据库 进行词频统计
以及词频大小排序
解决方案
一、实验目的
设计并实现链表、栈、队列基本操作，为编译原理的学习打下基础。
二、实验要求
实现线性表、栈、队列创建、查找、插入、删除等基本操作。
三、实验步骤
1.参考数据结构
//结点
typedef struct{
struct Node *next;
struct Node *pre;
int data;
}Node;
//链表头结点
struct
{
struct Node* head;
struct Node* tail;
int count;
} List;
typedef List Stack;//使用链表作为堆栈的底层实现结构
typedef List Queue;//使用链表作为队列的底层实现结构
2.操作
2.1 表
2.1.1. 初始化
void list_init(struct List* root){
root->count = 0;
root->head = NULL;
root->tail = NULL;
}
2.1.2 插入
//插入到队列末尾
void push_back(Node* node, List* list){
if(list->head == NULL){//如果list头结点为空
list->head = list->tail = node;
node->next = node->pre = NULL;
list->count++;
return;
}
node->next = NULL;
node->pre = list->tail;
list->tail->next = node;
list->tail = node;
list->count++;
}
//插入到队列头部
void push_front(Node node, List list){
if(list->head == NULL){
list->head = list->tail = node;
node->next = node->pre = NULL;
list->count++;
return;
}
node->pre= NULL;
node->next = list->head;
list->head->pre = node;
list->head = node;
list->count++;
}
//插入到位置locate
int insert(Node *node, List *list, int locate){
//异常处理
if(locate > list->count || locate < 0)return -1;
//插入到头部
if(locate == 0){push_front(node, list); return 1;}
//插入到尾部
if(locate == list->count){push_back(node, list); return 1;}
//其它
int curValue = -1;
//查找并插入到cur的前面
Node *cur = list->head;
while(++curValue < locate)cur = cur->next;
//插入
node->next = cur;
node->pre = cur->pre;
cur->pre->next = node;
cur->pre = node;
//容量加1
list->count++;
return 1;
}
2.1.3 查找
Node* search(Node *node, List *list, int &locate){
//返回相应的头结点，并记录该结点的位置到locate中
locate = 0;
Node *cur = list->head;
while(cur != NULL && cur->data != node->data){cur = cur->next; locate++;}
if(cur == NULL) locate = - 1;
return cur;
}
2.1.4. 删除
int remove(List *list, int locate){
//异常
if(locate >= list->count || locate < 0)return -1;
if(locate == 0){//头部
Node *p = list->head->next;
free(list->head);
p->pre = NULL;
list->head = p;
} else if(locate == list->count - 1){//尾部
Node *p = list->tail->pre;
free(list->tail);
p->next = NULL;
list->tail = p;
} else{//其它
//查找并保存到cur中
int curValue = -1;
Node *cur = list->head;
while(++curValue < locate)cur = cur->next;
cur->pre->next = cur->next;
cur->next->pre = cur->pre;
free(cur);
}
list->count--;
return 1;
}
2.2. 堆栈操作
2.2.1 初始化
void stack_init(Stack *s){
list_init((List*)s);
}
2.2.2 压栈
void push(struct Node* node, Stack* s){
push_front(node, (List*)s);
}
2.2.3 出栈
Node* pop(Stack* s){
Node *node = s->head;
if(s->head->next != NULL)
s->head = s->head->next;
return node;
}
2.2.4 获得栈顶
Node* get_top_stack(struct Node* node, Stack* s){
return s->head;
}
2.3. 堆栈操作
2.3.1. 初始化队列
void queue_init(Queue *q){
list_init((List*)q);
}
2.3.2. 入队列
void in_queue(struct Node* node, Queue* q){
push_back(node, (List*)q);
}
2.3.3. 出队列
Node* out_queue(Queue* q){
Node *node = q->head;
if(q->head->next != NULL)
q->head = q->head->next;
return node;
}
2.3.4. 获得队列头
Node* get_first_queue(struct Node* node, Queue* q){
return q->head;
}
解决方案二：
可以利用上面的东西。。
解决方案三：
君乐雨兮启兰枝，君乐日兮林蔽日
君乐风兮兰帐起，君乐吾兮吾心噬
解决方案四：
考虑下快速排序qsort