需求:我要在map中判断是否存在key,存在则让key对应的value = value+1,否则设置<key,value=1>
代码实现方式如下:
ContainsKey
import java.util.HashMap; import java.util.Map; ... Map<String, Integer> freq = new HashMap<String, Integer>(); ... int count = freq.containsKey(word) ? freq.get(word) : 0; freq.put(word, count + 1);
TestForNull
import java.util.HashMap; import java.util.Map; ... Map<String, Integer> freq = new HashMap<String, Integer>(); ... Integer count = freq.get(word); if (count == null) { freq.put(word, 1); } else { freq.put(word, count + 1); }
AtomicLong
import java.util.concurrent.ConcurrentHashMap; import java.util.concurrent.ConcurrentMap; import java.util.concurrent.atomic.AtomicLong; ... final ConcurrentMap<String, AtomicLong> map = new ConcurrentHashMap<String, AtomicLong>(); ... map.putIfAbsent(word, new AtomicLong(0)); map.get(word).incrementAndGet();
Trove
import gnu.trove.TObjectIntHashMap; ... TObjectIntHashMap<String> freq = new TObjectIntHashMap<String>(); ... freq.adjustOrPutValue(word, 1, 1);
MutableInt
import java.util.HashMap; import java.util.Map; ... class MutableInt { int value = 1; // note that we start at 1 since we're counting public void increment () { ++value; } public int get () { return value; } } ... Map<String, MutableInt> freq = new HashMap<String, MutableInt>(); ... MutableInt count = freq.get(word); if (count == null) { freq.put(word, new MutableInt()); } else { count.increment(); }
还有Java8的方法:
Map.merge(key, 1, Integer::sum) //如果key存在,就调用sum 1 到map的value,如果key不存在,put(key,1)
时间方面,执行1000次数,如下所示
time, ms kolobokeCompile 18.8 koloboke 19.8 trove 20.8 fastutil 22.7 mutableInt 24.3 atomicInteger 25.3 eclipse 26.9 hashMap 28.0 hppc 33.6 hppcRt 36.5