mcrwayfun
11835次阅读 2018-06-21
关注
1. 分析hashMap.containsKey
hashMap.containsKey(value)的时间复杂度为什么是O(1)呢？这个就要来看一下源码了
/**
* Returns <tt>true</tt> if this map contains a mapping for the
* specified key.
*
* @param key The key whose presence in this map is to be tested
* @return <tt>true</tt> if this map contains a mapping for the specified
* key.
*/
public boolean containsKey(Object key) {
return getNode(hash(key), key) != null;
}
1
2
3
4
5
6
7
8
9
10
11
调用了getNode(hash(key), key)方法，参数分别为key的hash值，key。再来看下这个方法的实现
/**
* Implements Map.get and related methods
*
* @param hash hash for key
* @param key the key
* @return the node, or null if none
*/
final Node<K,V> getNode(int hash, Object key) {
Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
if ((tab = table) != null && (n = tab.length) > 0 &&
(first = tab[(n - 1) & hash]) != null) {
// 直接命中
if (first.hash == hash && // always check first node
((k = first.key) == key || (key != null && key.equals(k))))
return first;
// 未命中
if ((e = first.next) != null) {
if (first instanceof TreeNode)
return ((TreeNode<K,V>)first).getTreeNode(hash, key);
do {
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
} while ((e = e.next) != null);
}
}
return null;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
可以看到返回的是first(默认check first node)，算法中带n的可能影响时间复杂度的便是：
first = tab[(n - 1) & hash]) != null
1
为了探究为什么是O(1)，这里就要理解
- Node<K,V> first是什么
- Node<K,V>[] tab是什么
Node<K,V> first：是一个单向链表结点，包含了hash，key，value和指向下个结点的指针
/**
* Basic hash bin node, used for most entries. (See below for
* TreeNode subclass, and in LinkedHashMap for its Entry subclass.)
*/
static class Node<K,V> implements Map.Entry<K,V> {
final int hash;
final K key;
V value;
Node<K,V> next;
Node(int hash, K key, V value, Node<K,V> next) {
this.hash = hash;
this.key = key;
this.value = value;
this.next = next;
}
... get and set method ...
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Node<K,V>[] tab：是一个数组，值得注意的是，数组长度为2的倍数
/**
* The table, initialized on first use, and resized as
* necessary. When allocated, length is always a power of two.
* (We also tolerate length zero in some operations to allow
* bootstrapping mechanics that are currently not needed.)
*/
transient Node<K,V>[] table;
1
2
3
4
5
6
7
所以tab[(n - 1) & hash]执行了如下操作：
1. 指针first指向那一行数组的引用（那一行数组是通过table下标范围n-1和key的hash值计算出来的），若命中，则通过下标访问数组，时间复杂度为O(1)
2. 如果没有直接命中（key进行hash时，产生相同的位运算值），存储方式变为红黑树，那么遍历树的时间复杂度为O(n)
2. 总结
综上，hashMap.containsKey(value)最好情况便是O(1)，最坏情况是O(n)
文章最后发布于: 2018-06-21
版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。
本文链接：https://blog.csdn.net/qingtian_1993/article/details/80763381
网友评论