以下是根據我對您疑問的理解編寫的示例程序。它會查找並顯示重複鍵以及重複值及其重複頻率。邏輯是:爲了找到頻率,將來自map1和map2的重複元素作爲參數傳遞給countFrequency()方法,其依次將返回重複次數。
package com.rahul.stackoverflow;
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
public class DuplicateValHashMap {
public static void main(String[] args) {
Map<String, Integer> map1 = new HashMap<String, Integer>();
Map<String, Integer> map2 = new HashMap<String, Integer>();
map1.put("A", 1);
map1.put("B", 2);
map1.put("C", 3);
map1.put("D", 4);
map1.put("E", 5);
map2.put("A", 1);
map2.put("F", 2);
map2.put("G", 1);
map2.put("H", 3);
map2.put("B", 2);
for(Entry<String, Integer> entrySet : map1.entrySet()){
if(map2.containsKey(entrySet.getKey())){
System.out.println("Map2 contains keys of map1.");
System.out.println("Duplicate keys are : " + entrySet.getKey());
}
if(map2.containsValue(entrySet.getValue())){
System.out.println("Map2 contains values of map1.");
System.out.println("Duplicate values are : " + entrySet.getValue()+
" which is repeated " + countFrequency(map2, entrySet.getValue())+ " times.");
}
}
}
public static int countFrequency(Map<String, Integer> map, Integer value){
int count = 0;
for(Entry<String, Integer> entrySet : map.entrySet()){
if(value == entrySet.getValue()){
count++;
}
}
return count;
}
}
我能想到的唯一方法是創建重複鍵和重複值的計數,累加計之間的一個映射,每當重複發現 –
你想要的是在這兩個地圖按鍵的數量,是那對嗎? – nandsito