[英]Cannot count number in ArrayList using java
I am using the List as shown below:- 我正在使用列表,如下所示:
List<Integer> integerlist = new ArrayList<Integer>();
integerlist=primeFactors(numberToBERadical);
The primeFactors method return me some number when i sysOut as:- 当我sysOut时,primeFactors方法返回给我一些数字:-
System.out.println("integer list=="+integerlist);
integer list==[2, 2, 2, 3, 5]
My Problem is i want to count how many time that each number appears so that i can show Like this:- 我的问题是我想计算每个数字出现多少次,以便我可以像这样显示:
2^3 * 3 * 5
You need to associate each factor with the number of times it occurs (the exponent). 您需要将每个因子与其发生的次数(指数)相关联。
To do that, a better data structure than a List would be a Map<Integer,Integer>
. 为此,比List更好的数据结构是
Map<Integer,Integer>
。
Let's assume you still get a List
from your primeFactors(numberToBERadical);
假设您仍然从
primeFactors(numberToBERadical);
获取List
primeFactors(numberToBERadical);
(assuming you can't change it or don't want to for some reason). (假设您无法更改它或出于某种原因而不想更改)。
You can do something like: 您可以执行以下操作:
public static void main(String[] args) {
List<Integer> list = new ArrayList<>();
list.add(2);
list.add(2);
list.add(2);
list.add(3);
list.add(3);
list.add(5);
Map<Integer, Integer> factors = new HashMap<>();
for(Integer fact: list){
if(!factors.containsKey(fact)) { factors.put(fact, 1);}
else {
Integer value = factors.get(fact);
factors.put(fact, ++value);
}
}
System.out.println(factors);
}
Prints: {2=3, 3=2, 5=1}
打印:
{2=3, 3=2, 5=1}
If the array is sorted, you can count them like: 如果数组已排序,则可以像这样对它们进行计数:
int count = 0;
int last = 0; // zero is not possible for prime-factors
for(int i=0;i<list.size();i++) {
if(i == 0 || last == list.get(i)) count++;
else {
if(last > list.get(0)) {
System.out.print(" * ");
}
System.out.print(last + (count > 1 ? "^" + count: ""));
count = 1;
}
last = list.get(i);
}
//print last sequence.
if(last > 0) {
if(last > list.get(0)) {
System.out.print(" * ");
}
System.out.print(last + (count > 1 ? "^" + count: ""));
}
One possible, but not the simplest solution: 一种可能,但不是最简单的解决方案:
public Map<Integer, AtomicInteger> primeFactors(List<Integer integerList) {
final Map<Integer, AtomicInteger> count = new HashMap<Integer, AtomicInteger>();
for (final Integer number : integerlist) {
if (count.containsKey(number)) {
count.get(number).incrementAndGet();
} else {
count.put(number, new AtomicInteger(1));
}
}
return count;
}
You need to use a Map<Integer, Integer>
which you can override the put
method in in order to count the instances of each Integer
. 您需要使用
Map<Integer, Integer>
,您可以重写put
方法以便计算每个Integer
的实例。
final Map<Integer, Integer> myStringMap = new HashMap<>(){
@override
public String put(final Integer key, final Integer value) {
if(contains(key)) {
return put(key, get(key) + 1);
} else {
return put(key, 1);
}
}
};
You could even use a TreeMap
to have the prime factors sorted by size. 您甚至可以使用
TreeMap
将主要因子按大小排序。 I answered a very similar question here . 我在这里回答了一个非常类似的问题。 Simply loop over your
List
and dump it into the map 只需遍历您的
List
并将其转储到地图中
for(final Integer integer : myList) {
myCountingMap.put(integer, 1);
}
Even better would be to change your factorising method to return a Map
to begin with. 更好的办法是更改分解方法以返回
Map
。
List<Integer> inputList = new ArrayList<Integer>();
inputList.add(2);
inputList.add(2);
inputList.add(2);
inputList.add(3);
inputList.add(3);
inputList.add(4);
Map<Integer, Integer> resultMap = new HashMap<Integer, Integer>();
boolean flag = false;
for(int val : inputList)
{
if(resultMap.get(val) == null)
{
resultMap.put(val, 1);
}
else
{
resultMap.put(val, (resultMap.get(val).intValue())+1);
}
}
for(int key : resultMap.keySet())
{
if(resultMap.get(key) == 1)
{
if(!flag)
{
System.out.print(key);
flag = true;
}
else
{
System.out.print("*"+key);
}
}
else
{
if(!flag)
{
System.out.print(key+"^"+resultMap.get(key));
flag = true;
}
else
{
System.out.print("*"+key+"^"+resultMap.get(key));
}
}
}
System.out.println();
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