我一直在研究GeeksforGeeks的排列問題。這裏是挑戰的鏈接:http://www.practice.geeksforgeeks.org/problem-page.php?pid=702Java排列挑戰是緩慢的
挑戰是採取一系列的數字和這些數字的每個可能的順序在2或3的組中,檢查數字的總和是否可以被3整除。的節目的結束打印出一個例子是,通過3
整除組的數量... INT []數組= {1,2,3}應該打印出8.
以下是我用於這個挑戰的代碼。代碼有效,但速度很慢。運行時需要低於1.272s,所以我怎樣才能使這個代碼更快?或者保存它執行很多行?
public static void PG2(int[] array, int l, int r, Counter count){
int newR = r - 1;
int i;
if(l == 2){
String number = String.valueOf(array[0]);
String number2 = String.valueOf(array[1]);
number = number.concat(number2);
int aNumber = Integer.parseInt(number);
count.divBy3(aNumber);
} else {
int temp;
int temp2;
for(i = l; i < r; i++){
temp = array[l];
array[l] = array[i];
array[i] = temp;
PG2(array, l+1, r, count);
temp2 = array[l];
array[l] = array[i];
array[i] = temp2;
}
}
}
public static void PG3(int[] array, int l, int r, Counter count){
int newR = r - 1;
int i;
if(l == 3){
String number = String.valueOf(array[0]);
String number2 = String.valueOf(array[1]);
String number3 = String.valueOf(array[2]);
number = number.concat(number2);
number = number.concat(number3);
int aNumber = Integer.parseInt(number);
count.divBy3(aNumber);
} else {
int temp;
int temp2;
for(i = l; i < r; i++){
temp = array[l];
array[l] = array[i];
array[i] = temp;
PG3(array, l+1, r, count);
temp2 = array[l];
array[l] = array[i];
array[i] = temp2;
}
}
}
public static void main(String[] args) throws Exception {
BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
String t = input.readLine();
int T = Integer.parseInt(t);
while(T > 0){
String arrayS = input.readLine();
int ArrayS = Integer.parseInt(arrayS);
int[] newArray = new int[ArrayS];
String arrayElements = input.readLine();
String[] ArrayElements = arrayElements.trim().split("\\s+");
for(int i = 0; i < ArrayS; i++){
int num = Integer.parseInt(ArrayElements[i]);
newArray[i] = num;
}
int total = 0;
Counter count = new Counter();
PG2(newArray, 0, ArrayS, count);
PG3(newArray, 0, ArrayS, count);
System.out.println(count.counter);
T--;
}
}
這裏是計數器類:
public class Counter {
public int counter;
public void divBy3(int number){
int total = 0;
while(number > 0){
total += number % 10;
number = number/10;
}
if(total % 3 == 0){
this.counter++;
}
}
}
什麼代碼需要1.2秒? –
如果您可以使用表達式,請使用Java 8流API。 請參閱:https://docs.oracle.com/javase/tutorial/java/javaOO/lambdaexpressions.html – amitmah