Java中遍历大容量map的正确方法

import java.util.HashMap;
import java.util.Map;

public class MapTest {
	static long MAX_LONG = 1000000L;
	static int TIMES = 10;
	static Map<String, String> map1 = new HashMap<String, String>();
	static Map<String, String> map2 = new HashMap<String, String>();
	static {
		for (long i = 0; i < MAX_LONG; i++) {
			map1.put("1" + i, "abc" + i);
			map2.put("2" + i, "def" + i);
		}
	}

	public static void main(String[] args) {

		String valueStr = ""; 
		String keyStr = "";
		long start, end;
		double totalMs;



		totalMs = 0;
		for (int i = 0; i < TIMES; i++) {

			// 通过Map.keySet遍历key和value
			start = System.currentTimeMillis();
			for (String key : map1.keySet()) {
				keyStr = key;
				valueStr = map1.get(key);
			}
			end = System.currentTimeMillis();
			System.out.println("通过Map.keySet遍历key和value耗时 " + (end - start)
					+ " ms ");
			totalMs += (end - start);
		}
		System.out.println("Times : " + TIMES + ", totalMs : " + totalMs
				+ "ms, average :" + totalMs / TIMES + "ms");
		
		
		totalMs = 0;
		for (int i = 0; i < TIMES; i++) {
			// 通过Map.entrySet遍历key和value
			start = System.currentTimeMillis();
			for (Map.Entry<String, String> entry : map2.entrySet()) {
				keyStr = entry.getKey();
				valueStr = entry.getValue();
			}
			end = System.currentTimeMillis();
			System.out.println("通过Map.entrySet遍历key和value耗时 " + (end - start)
					+ " ms ");
			totalMs += (end - start);
		}
		System.out.println("Times : " + TIMES + ", totalMs : " + totalMs
				+ "ms, average :" + totalMs / TIMES + "ms");

	}

}

通过Map.keySet遍历key和value耗时 186 ms 
通过Map.keySet遍历key和value耗时 189 ms 
通过Map.keySet遍历key和value耗时 87 ms 
通过Map.keySet遍历key和value耗时 89 ms 
通过Map.keySet遍历key和value耗时 84 ms 
通过Map.keySet遍历key和value耗时 92 ms 
通过Map.keySet遍历key和value耗时 85 ms 
通过Map.keySet遍历key和value耗时 94 ms 
通过Map.keySet遍历key和value耗时 89 ms 
通过Map.keySet遍历key和value耗时 91 ms 
Times : 10, totalMs : 1086.0ms, average :108.6ms
通过Map.entrySet遍历key和value耗时 112 ms 
通过Map.entrySet遍历key和value耗时 98 ms 
通过Map.entrySet遍历key和value耗时 71 ms 
通过Map.entrySet遍历key和value耗时 65 ms 
通过Map.entrySet遍历key和value耗时 65 ms 
通过Map.entrySet遍历key和value耗时 64 ms 
通过Map.entrySet遍历key和value耗时 64 ms 
通过Map.entrySet遍历key和value耗时 65 ms 
通过Map.entrySet遍历key和value耗时 65 ms 
通过Map.entrySet遍历key和value耗时 65 ms 
Times : 10, totalMs : 734.0ms, average :73.4ms

map.get(key))

时执行的hash操作

/**
* Returns the value to which the specified key is mapped,
* or {@code null} if this map contains no mapping for the key.
*
* <p>More formally, if this map contains a mapping from a key
* {@code k} to a value {@code v} such that {@code (key==null ? k==null :
* key.equals(k))}, then this method returns {@code v}; otherwise
* it returns {@code null}.  (There can be at most one such mapping.)
*
* <p>A return value of {@code null} does not <i>necessarily</i>
* indicate that the map contains no mapping for the key; it's also
* possible that the map explicitly maps the key to {@code null}.
* The {@link #containsKey containsKey} operation may be used to
* distinguish these two cases.
*
* @see #put(Object, Object)
*/
public V get(Object key) {
if (key == null)
    return getForNullKey();
int hash = hash(key.hashCode());
for (Entry<K,V> e = table[indexFor(hash, table.length)];
     e != null;
     e = e.next) {
    Object k;
    if (e.hash == hash && ((k = e.key) == key || key.equals(k)))
	return e.value;
}
return null;
}
    

无奇不有地