Question
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;
}
}
原文地址:https://www.cnblogs.com/okokabcd/p/9374563.html
时间: 2024-11-09 07:04:55