我有两个我需要合并的哈希图。
MAP 1 - [[LOCATION: United Kingdom], [PERSON: Alfred Theodore MacConkey], [ORGANIZATION: Royal Commission, United]]
MAP 2 -{LOCATION=[United Kingdom], ORGANIZATION=[Royal Commission], PERSON=[Alfred Theodore MacConkey]}
如果我创建地图3并执行addall()一些值会被覆盖。就像组织的情况下,我只会得到皇家委员会,而曼联则被覆盖。我为两者编写了合并代码,但得到了无效的指针例外。我只想知道这种方法是否正确。我会调试并弄清楚为什么我会得到例外。
public static LinkedHashMap<String,Vector<String>> merge(HashMap<String, Vector<String>> a, HashMap<String, Vector<String>> b) {
LinkedHashMap<String,Vector<String>> c = new LinkedHashMap<String,Vector<String>>();
Set<Entry<String,Vector<String>>> entriesA = a.entrySet();
Set<Entry<String,Vector<String>>> entriesB = b.entrySet();
for (Map.Entry<String, Vector<String>> entry : entriesA ) {
Vector<String> aValues = a.get(entry.getValue());
String aKey = entry.getKey();
Vector<String> allValues = entriesA.contains(aKey) ? a.get(aKey) : new Vector<String>();
allValues.addAll(aValues);
c.put(aKey, allValues);
}
for (Map.Entry<String, Vector<String>> entry : entriesB ) {
Vector<String> bValues = b.get(entry.getValue());
String bKey = entry.getKey();
if(c.containsKey(bKey) && c.get(bKey).equals(bValues) ) {
continue;
}
else if(c.containsKey(bKey) && !(c.get(bKey).equals(bValues))) {
c.put(bKey, bValues);
}
}
return c;
}
此行:
Vector<String> aValues = a.get(entry.getValue());
应该是:
Vector<String> aValues = entry.getValue();
更新:
哦!bvalues
也是如此更新2:
还有另一个问题:Entriesa.Contains(Akey)应为A.Contains(Akey)
更新3:
尝试这样的事情:
LinkedHashMap<String, Vector<String>> c = new LinkedHashMap<String, Vector<String>>();
for (Map.Entry<String, Vector<String>> entry : a.entrySet()) {
Vector<String> aValues = entry.getValue();
String aKey = entry.getKey();
c.put(aKey, new Vector<String>(aValues));
}
for (Map.Entry<String, Vector<String>> entry : b.entrySet()) {
Vector<String> bValues = entry.getValue();
String bKey = entry.getKey();
Vector<String> cValues = c.get(bKey);
if (cValues == null) {
c.put(bKey, new Vector<String>(bValues));
} else {
cValues.addAll(bValues);
}
}
return c;
更新4:
要避免重复值,请替换行:
cValues.addAll(bValues);
with:
Set<String> values = new HashSet<String>(cValues);
values.addAll(bValues);
cValues.clear();
cValues.addAll(values);
这只会处理合并创建的重复项,而不是已经存在的重复项。