我目前正在开发一个学校项目，该项目将解析包含不同移动指令的文本文件。然后将这些指令放在一个树结构中，该树结构总是用新指令向右括起来，在左边用每条指令可以走多少。
它看起来像这样：
FORW
/ \
2 LEFT
/ \
90 Rep
/ \
FORW FORW
/ /
4 2
无论如何，如果没有详细了解我的程序细节，我的问题就是如何改进程序的这一部分以使其变得更快？
正如你所看到的，我已经尝试过使用stringbuilder来加速这个过程，但它并没有太大的不同，在这一点上我有点想法，所以任何帮助都会非常感激。
这个程序经过了很多测试用例的测试，如果它们中的任何一个需要超过7秒，它就会失败，现在就是这种情况。
import java.util.ArrayList;
// Ett syntaxträd
public class CreateLines{
private boolean penDown; // 1 om pennan är nere 0 om inte.
private double x1, x2, y1, y2; // x,y-koordinat
private int degrees; // vinkel v
private String color; // nuvarande färg
private double piDegrees;
private double decimals;
private ArrayList<String> result;
public CreateLines() {
penDown = false;
x1 = 0;
x2 = 0;
y1 = 0;
y2 = 0;
degrees = 0;
color = "#0000FF";
// For optimization.
decimals = 100000;
}
public StringBuilder treeSearch (Operation tree) {
// Some variables we use down here:
double x1 = this.x1;
double y1 = this.y1;
double x2;
double y2;
int numberNode;
StringBuilder str = new StringBuilder();
switch (tree.operation) {
case FORW:
piDegrees = Math.PI*degrees/180;
numberNode = tree.evaluate();
x2 = x1 + (numberNode * Math.cos(piDegrees));
y2 = y1 + (numberNode * Math.sin(piDegrees));
x2 = (double)Math.rint(x2 * decimals) / decimals;
y2 = (double)Math.rint(y2 * decimals) / decimals;
this.x1 = x2;
this.x2 = x2;
this.y1 = y2;
this.y2 = y2;
// If penDown then we print otherwise not.
if (penDown){
str.append(color + " " + x1 + " " + y1 + " " + x2 + " " + y2 + "\n");
if(tree.right == null){
return str;
}
return str.append(treeSearch((Operation) tree.right));
}
else{
if(tree.right == null){
return str;
}
return treeSearch((Operation) tree.right);
}
case BACK:
piDegrees = Math.PI*degrees/180;
numberNode = tree.evaluate();
x2 = x1 - (numberNode * Math.cos(piDegrees));
y2 = y1 - (numberNode * Math.sin(piDegrees));
x2 = (double)Math.rint(x2 * decimals) / decimals;
y2 = (double)Math.rint(y2 * decimals) / decimals;
this.x1 = x2;
this.x2 = x2;
this.y1 = y2;
this.y2 = y2;
// If penDown then we print otherwise not.
if (penDown){
str.append(color + " " + x1 + " " + y1 + " " + x2 + " " + y2 + "\n");
if(tree.right == null){
return str;
}
return str.append(treeSearch((Operation) tree.right));
}
else{
if(tree.right == null){
return str;
}
return treeSearch((Operation) tree.right);
}
case LEFT:
numberNode = tree.evaluate();
this.degrees = degrees+numberNode;
if (penDown){
if(tree.right == null){
return str;
}
return treeSearch((Operation) tree.right);
}
else{
if(tree.right == null){
return str;
}
return treeSearch((Operation) tree.right);
}
case RIGHT:
numberNode = tree.evaluate();
this.degrees = degrees-numberNode;
if (penDown){
if(tree.right == null){
return str;
}
return treeSearch((Operation) tree.right);
}
else{
if(tree.right == null){
return str;
}
return treeSearch((Operation) tree.right);
}
case DOWN:
this.penDown = true;
if(tree.right == null){
return str;
}
return treeSearch((Operation) tree.right);
case UP:
this.penDown = false;
if(tree.right == null){
return str;
}
return treeSearch((Operation) tree.right);
case COLOR:
this.color = tree.color.toUpperCase();
if (penDown){
if(tree.right == null){
return str;
}
return treeSearch((Operation) tree.right);
}
else{
if(tree.right == null){
return str;
}
return treeSearch((Operation) tree.right);
}
case REP:
// if we got a rep instruction to the left we
if(tree.right == null){
for(int i = 0; i < tree.rep; i++){
str.append(treeSearch((Operation) tree.left));
}
return str;
}
else {
for(int i = 0; i < tree.rep; i++){
str.append(treeSearch((Operation) tree.left));
}
return str.append(treeSearch((Operation)tree.right));
}
}
assert false; // borde aldrig kunna hända
return null;
}
}
答案 0 :(得分：0)
经过一番思考后，Tedil和OeterLawrey指出问题是我通过为每次调用treeSearch创建一个新的StringBuilder而做了一个很大的错误，因此我的程序非常慢。而不是返回每个Stringbuilder我改为创建一个变量str，然后用一个新方法将它返回给主程序，通过这样做我解决了问题。
感谢帮帮！