就输出而言,我的程序运行良好,但是对于我的一些测试用例来说,找到答案的时间太长(有时需要18秒)。我想知道如何改善代码的性能。
我的代码做什么:
这是对Pebble Solitaire的看法。用户输入n个游戏,然后输入长度为23的字符串,其中仅包含“ o”(卵石)和“-”(空白空间)的组合。如果有两个相邻的小卵石,并且在两侧都有一个空白空间,即(oo-或-oo),则将中间的小卵石移开,并将其他两块卵互换,例如“ oo-”将变成“-哦
我当前的方法几乎是一种详尽的方法,它尝试了所有可能的移动,并以最小的卵石数量产生了移动集。
我想知道如何在不使其成为多线程的情况下改进该解决方案。
这是我所拥有的:
package Pebble;
import java.util.Scanner;
public class PebbleSolitaire {
public static void main(String[] args){
Scanner input = new Scanner(System.in);
int numOfGames = Integer.parseInt(input.nextLine());
while (numOfGames > 0){
char[] values = input.nextLine().toCharArray();
long startTime = System.nanoTime();
System.out.println(solve(values));
System.out.println("Time to finish in ms: " + (System.nanoTime() - startTime) / 1000000);
numOfGames--;
}
input.close();
}
private static int solve(char[] game){
if(game != null && game.length == 0){
return -1;
}
int result = 0;
for (int i = 0; i < game.length; i++){
if(game[i] == 'o'){
result++;
}
}
//print(game);
for (int i = 0; i < game.length; i++ ){
char[] temp = new char[game.length];
copyArray(temp, game);
if (i-2 >= 0 && temp[i] == '-' && temp[i-2] == 'o' && temp[i-1] == 'o'){//move pebble forwards
temp[i-1] = temp[i-2] = '-';
temp[i] = 'o';
result = Math.min(result, solve(temp));
}
copyArray(temp, game);
if(i+2 < temp.length && temp[i] == '-' && temp[i+1] == 'o' && temp[i+2] == 'o'){//move pebble backwards
temp[i+1] = temp[i+2] = '-';
temp[i] = 'o';
result = Math.min(result, solve(temp));
}
}
return result;
}
private static void copyArray(char[] copy, char[] og){
for(int x = 0; x < copy.length; x++){
copy[x] = og[x];
}
}
private static void print(char[] c){
for(char ch: c){
System.out.print(ch);
}
System.out.println();
}
}
我的样本输入和输出:
2
-o----ooo----o----ooo--
6
Time to finish in ms: 0
oooooooooo-ooooooooooo-
4
Time to finish in ms: 18149
编辑:进行此完全迭代会大大提高性能吗?
最佳答案
也许您可以改善这一方面:
for (int i = 0; i < game.length; i++ ){
char[] temp = new char[game.length];
copyArray(temp, game);
if (i-2 >= 0 && temp[i] == '-' && temp[i-2] == 'o' && temp[i-1] == 'o'){//move pebble forwards
temp[i-1] = temp[i-2] = '-';
temp[i] = 'o';
result = Math.min(result, solve(temp));
}
copyArray(temp, game);
if(i+2 < temp.length && temp[i] == '-' && temp[i+1] == 'o' && temp[i+2] == 'o'){//move pebble backwards
temp[i+1] = temp[i+2] = '-';
temp[i] = 'o';
result = Math.min(result, solve(temp));
}
}
至:
for (int i = 0; i < game.length; i++ ){
char[] temp = null;
if (i-2 >= 0 && game[i] == '-' && game[i-2] == 'o' && game[i-1] == 'o'){//move pebble forwards
temp = new char[game.length];
copyArray(temp, game);
temp[i-1] = temp[i-2] = '-';
temp[i] = 'o';
result = Math.min(result, solve(temp));
}
if(i+2 < game.length && game[i] == '-' && game[i+1] == 'o' && game[i+2] == 'o'){//move pebble backwards
if(temp == null) temp = new char[game.length];
copyArray(temp, game);
temp[i+1] = temp[i+2] = '-';
temp[i] = 'o';
result = Math.min(result, solve(temp));
}
}
基本上,仅创建和“ copyArray(temp,game);”严格必要时。