实现一个简易的HashMap

实现一个键的类型为int,值的类型为int的HashMap
输入一个T,表示操作次数;
之后每行接一个操作,可以包括插入、删除、修改、查询、清空、判断是否有这个键;
因为是刚学完随手敲的,所以功能粗糙。插入不考虑键已经存在的情况。删除、修改、查询不考虑键不存在的情况;

#include "bits/stdc++.h"
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
// 因为int的数据范围为1 << 32 个数。所以把哈希表开到1 << 16从内存和效率上讲比较折中
const int SIZE = 1 << 16;
struct Map {
    int key;
    int value;
    Map* next;
}*hashtable[SIZE];
int hashCode(int k) {
    int code = k % SIZE;
    return code < 0 ? -code : code;
}
Map* init(int k, int v, Map* next) {
    Map* point = (Map*)malloc(sizeof(Map));
    point->key = k;
    point->value = v;
    point->next = next;
    return point;
}
void insert(int k, int v) {
    int id = hashCode(k);
    hashtable[id] = init(k, v, hashtable[id]);
}
void update(int k, int v) {
    int id = hashCode(k);
    Map* point = hashtable[id];
    while (point != NULL) {
        if (point->key == k) {
            point->value = v;
            return;
        }
        point = point->next;
    }
}
int query(int k) {
    int id = hashCode(k);
    Map* point = hashtable[id];
    while (point != NULL) {
        if (point->key == k) {
            return point->value;
        }
        point = point->next;
    }
}
bool hasKey(int k) {
    int id = hashCode(k);
    Map* point = hashtable[id];
    while (point != NULL) {
        if (point->key == k) {
            return true;
        }
        point = point->next;
    }
    return false;
}
void erase(int k) {
    int id = hashCode(k);
    Map* pre = hashtable[id];
    if (pre == NULL){
        return;
    }
    if (pre->key == k) {
        hashtable[id] = NULL;
        return;
    }
    Map* now = pre->next;
    while (now != NULL) {
        if (now->key == k) {
            pre->next = now->next;
            free(now);
            return;
        }
        pre = now;
        now = now->next;
    }
}
void clear() {
    for (int i = 0; i < SIZE; i++) {
        Map* point = hashtable[i];
        Map* next;
        while (point != NULL) {
            next = point->next;
            free(point);
            point = next;
        }
        hashtable[i]=NULL;
    }
}
int main() {
    int T, k, v;
    char op[10];
    scanf("%d", &T);
    while (T--) {
        scanf("%s", op);
        if (strcmp("insert", op) == 0) {
            scanf("%d%d", &k, &v);
            insert(k, v);
        } else if (strcmp("update", op) == 0) {
            scanf("%d%d", &k, &v);
            update(k, v);
        } else if (strcmp("query", op) == 0) {
            scanf("%d", &k);
            printf("%d\n", query(k));
        } else if (strcmp("hasKey", op) == 0) {
            scanf("%d", &k);
            puts(hasKey(k) ? "Yes" : "No");
        } else if (strcmp("erase", op) == 0) {
            scanf("%d", &k);
            erase(k);
        } else if (strcmp("clear", op) == 0) {
            clear();
        }
    }
    return 0;
}

 

posted @ 2019-01-01 14:12  Jathon-cnblogs  阅读(457)  评论(0编辑  收藏  举报