数据结构与算法之散列表(拉链法解决散列冲突)
自己手撸了一下拉链法解决散列冲突的算法,代码如下:#include <iostream>#include <vector>#define N 1000using namespace std;typedef struct ChainNodes {int data;ChainNodes* Next;}ChainNodes;void CreatHashTable(vector&l
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自己手撸了一下拉链法解决散列冲突的算法,代码如下:
#include <iostream>
#include <vector>
#define N 1000
using namespace std;
typedef struct ChainNodes {
int data;
ChainNodes* Next;
}ChainNodes;
void CreatHashTable(vector<ChainNodes*>& try01) {
//ChainNodes** try01 = new ChainNodes * [10];//try应该是关键字
for (int i = 0; i < try01.size(); i++) {
try01[i] = NULL;
}
cout << "请输入您要储存的数字" << endl;
int item, i = 0;
char t;
while (cin >> item) {
//“=”在C语言中优先级中最低;
i = item % 7;
if (try01[i] == NULL) {
try01[i] = new ChainNodes;
try01[i]->data = item;
try01[i]->Next = NULL;
}
else {
ChainNodes* pre = try01[i];
ChainNodes* cur = try01[i]->Next;
while (cur != NULL) {
pre = cur;//“=”在C语言中优先级中最低;
cur = cur->Next;
}
pre->Next = new ChainNodes;/*这段代码的目的是进行链表的尾插法;但是当try指向空节点时,没有一个临时指针保存try01的前继节点,导致try01
指向一片和链表中断的空间,之后再把temp赋给try01,则原先try01【i】所指的空间失去连接,导致内存泄漏*/
pre->Next->data = item;
pre->Next->Next = NULL;
};
if ((t = cin.get()) == '\n') break;
}
};
void QueryNum(vector<ChainNodes*>& try01) {
cout << "请输入您要查询的数" << endl;
int tem;
cin >> tem;
if (try01[tem % 7] == NULL) {
cout << "您要查询的数不存在" << endl;
}
else {
ChainNodes* temp = try01[tem % 7];
int count = 0;
while (temp != NULL) {
count++;//
if (temp->data == tem) {
cout << "您找的数位于散列表第" << tem % 7 << "个表的第" << count << "个链表中" << endl;
break;
}
temp = temp->Next;
}
if (temp == NULL) {
cout << "您要查询的数不存在" << endl;
}
}
}
void DeleteHashTable(vector<ChainNodes*>& try01) {
for (int i = 0; i < try01.size(); i++) {
if (try01[i] == NULL) {
cout << "第"<<i<<"位链表为空,无需释放。" << endl;
}
else {
ChainNodes* temp = try01[i];
int count = 0;
while (temp != NULL) {
ChainNodes* t = temp;
cout << t->data << endl;
temp = temp->Next;
delete t;
}
if (temp == NULL) {
try01[i] = NULL;
cout << "第"<<i<<"位链表已经释放完毕" << endl;
}
}
}
}
int main()
{
vector<ChainNodes*> try01(10);
CreatHashTable(try01);
QueryNum(try01);
DeleteHashTable(try01);
return 0;
}
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