模拟散列表代替map(java)

 

//拉链法

import java.io.*;
import java.util.Scanner;


public class Main {
    //N要取大于100000的最小素数
    static int N = 100003,idx;
    static int[] h = new int[N];    //用于存放输入的数据
    static int[] e = new int[N];    //用来存放节点的值
    static int[] ne = new int[N];    //用来存放节点的next指针
    
    public static void add(int x){
        int k = (x % N + N) % N;    //k为要插入的位置
        e[idx] = x; //先将x这个值存下来
        ne[idx] = h[k]; //头节点是h[k],先让idx指向头节点
        h[k] = idx++;   //再把头节点指向新节点idx,idx用完后指向下一位
    }

    public static boolean find(int x){
        int k = (x % N + N) % N;
        for (int i = h[k]; i != -1; i = ne[i]) {
            if (e[i] == x) return true;
        }
        return false;
    }
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        idx = 0;
        for (int i = 0; i < N; i++) {
            h[i] = -1;
        }
        while (n-- > 0){
            String x = sc.next();
            int nums = sc.nextInt();
            if (x.equals("I")){
                add(nums);
            } else {
                if (find(nums)) System.out.println("Yes");
                else System.out.println("No");
            }
        }
    }
}





//开放寻址法
package Test;

import java.io.*;
import java.util.Scanner;


public class Main {
    //N要取大于100000的最小素数
    static int N = 100003,idx;
    static int[] h = new int[N];    //用于存放输入的数据
    static int nulls = 0x3f3f3f3f;

    //找到x应该插入的位置
    public static int find(int x){
        int k = (x % N + N) % N;
        //h[k]已经被用过,但不是被x用的,需要让x往下走
        while (h[k] != nulls && h[k] != x){
            k++;
            if (k == N) k = 0;
        }
        return k;
    }
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        idx = 0;
        for (int i = 0; i < N; i++) {
            h[i] = 0x3f3f3f3f;
        }
        while (n-- > 0){
            String x = sc.next();
            int nums = sc.nextInt();
            
            //通过find函数找到x应该放的下标
            int k = find(nums);
            if (x.equals("I")){
                h[k] = nums;
            } else {
                //通过nulls判断该下标上是否有被赋值过
                if (h[k] != nulls) System.out.println("Yes");
                else System.out.println("No");
            }
        }
    }
}