706. Design HashMap - LeetCode

最新推荐文章于 2025-08-12 18:32:44 发布

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最新推荐文章于 2025-08-12 18:32:44 发布

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文章标签：数据结构与算法 java

原文链接：http://www.cnblogs.com/okokabcd/p/9374563.html

本文提供了一种使用数组实现简单HashMap的方法，并介绍了一个更高效的解决方案，利用哈希表和链表来实现put、get和remove操作。

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Question

706. Design HashMap

Solution

题目大意:构造一个hashmap

思路:讨个巧,只要求key是int,哈希函数选择f(x)=x,规定key最大为1000000,那构造一个1000000的数组

Java实现:

class MyHashMap {

    int[] table;
    
    /** Initialize your data structure here. */
    public MyHashMap() {
        table = new int[1000000];
        Arrays.fill(table, -1);
    }
    
    /** value will always be non-negative. */
    public void put(int key, int value) {
        table[key] = value;
    }
    
    /** Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key */
    public int get(int key) {
        return table[key];
    }
    
    /** Removes the mapping of the specified value key if this map contains a mapping for the key */
    public void remove(int key) {
        table[key] = -1;
    }
}

/**
 * Your MyHashMap object will be instantiated and called as such:
 * MyHashMap obj = new MyHashMap();
 * obj.put(key,value);
 * int param_2 = obj.get(key);
 * obj.remove(key);
 */

参考别人的

class MyHashMap {
    final Bucket[] buckets = new Bucket[10000];

    public void put(int key, int value) {
        int i = idx(key);
        if (buckets[i] == null)
            buckets[i] = new Bucket();
        ListNode prev = find(buckets[i], key);
        if (prev.next == null)
            prev.next = new ListNode(key, value);
        else prev.next.val = value;
    }

    public int get(int key) {
        int i = idx(key);
        if (buckets[i] == null)
            return -1;
        ListNode node = find(buckets[i], key);
        return node.next == null ? -1 : node.next.val;
    }

    public void remove(int key) {
        int i = idx(key);
        if (buckets[i] == null) return;
        ListNode prev = find(buckets[i], key);
        if (prev.next == null) return;
        prev.next = prev.next.next;
    }

    int idx(int key) { return Integer.hashCode(key) % buckets.length;}

    ListNode find(Bucket bucket, int key) {
        ListNode node = bucket.head, prev = null;
        while (node != null && node.key != key) {
            prev = node;
            node = node.next;
        }
        return prev;
    }
}

class Bucket {
    final ListNode head = new ListNode(-1, -1);
}

class ListNode {
    int key, val;
    ListNode next;

    ListNode(int key, int val) {
        this.key = key;
        this.val = val;
    }
}

转载于:https://www.cnblogs.com/okokabcd/p/9374563.html