面临一个问题,我需要将元素放在两个不同的 Map 中,然后遍历所有这些并将 if 条件放在最后一个 Map 的最后一个元素上。
在这里,我有我尝试过的代码:
private Map<Integer, String> testing1 = new TreeMap<>();
private Map<Integer, String> testing2 = new TreeMap<>();
testing1.put(1, "one");
testing1.put(2, "two");
testing2.put(3, "three");
testing2.put(4, "four");
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现在我有两个 for 循环,我想检查两个 for 循环现在是否都有最后一个元素:我尝试过 .size() 但这对 Map 没有帮助
for (Map.Entry<Integer,String> something: testing1.entrySet()) {
for (Map.Entry<Integer,String> something2: testing2.entrySet()) {
if(){
}
}
}
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我想,这对你真的很有用。
TreeMap 的lastKey方法。
Map<Integer, String> testing1 = new TreeMap<>();
Map<Integer, String> testing2 = new TreeMap<>();
testing1.put(1, "one");
testing1.put(2, "two");
testing2.put(3, "three");
testing2.put(4, "four");
int lastKey1 = ((TreeMap<Integer, String>) testing1).lastKey();
int lastKey2 = ((TreeMap<Integer, String>) testing2).lastKey();
for (Map.Entry<Integer,String> something: testing1.entrySet()) {
for (Map.Entry<Integer,String> something2: testing2.entrySet()) {
if(something.getKey() == lastKey1 && something2.getKey() == lastKey2){
// do your work
}
}
}
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