Acwing 840. 模拟散列表

题目描述

思路讲解

代码展示

拉链法：

#include 
#include 

using namespace std;

const int N = 100003;

int h[N], e[N], ne[N], idx;

void insert(int x) {
    int k = (x % N + N) % N;
    e[idx] = x;
    ne[idx] = h[k];
    h[k] = idx++;
}

bool find(int x) {
    int k = (x % N + N) % N;  //让余数变成正数
    for (int i = h[k]; i != -1; i = ne[i])
        if (e[i] == x)
            return true;

    return false;
}

int main() {
    int n;
    scanf("%d", &n);

    memset(h, -1, sizeof h);

    while (n--) {
        char op[2];
        int x;
        scanf("%s%d", op, &x);

        if (*op == 'I') insert(x);
        else {
            if (find(x)) puts("Yes");
            else puts("No");
        }
    }

    return 0;
}

开放寻址法：

#include 
#include 

using namespace std;

const int N = 200003, null = 0x3f3f3f3f;

int h[N];

int find(int x) {
    int t = (x % N + N) % N;
    while (h[t] != null && h[t] != x) {
        t++;
        if (t == N) t = 0;
    }
    return t;
}

int main() {
    memset(h, 0x3f, sizeof h);

    int n;
    scanf("%d", &n);

    while (n--) {
        char op[2];
        int x;
        scanf("%s%d", op, &x);
        if (*op == 'I') h[find(x)] = x;
        else {
            if (h[find(x)] == null) puts("No");
            else puts("Yes");
        }
    }

    return 0;
}