嗨,我想打破这个循环,在两个if语句都为true时返回坐标。但是循环永远不会结束。 我该如何解决?
public static String[] positionQuery(int dim, Scanner test_in) {
Scanner scanner = new Scanner(System.in);
System.out.println("Provide origin and destination coordinates.");
System.out.println("Enter two positions between A1-H8:");
while(true) {
String line = scanner.nextLine();
String[] coordinates = line.split(" ");
if(coordinates.length == 2) {
String origin = coordinates[0];
String dest = coordinates[1];
if(validCoordinate(origin, dim) && validCoordinate(dest,dim)) {
return coordinates;
}
}
System.out.println("ERROR: Please enter valid coordinate pair separated by space.");
}
}
我想我对validCoordinates有问题,因为我只是对validCoordinates有所了解,但找不到我做错的事情。
public static boolean validCoordinate(String coordinate, int dimension) {
boolean isValidCoordinate;
String [] alphabet = {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
int [] numbers = new int [dimension];
int one = 1;
for(int i = 0; i < dimension; i++){
numbers[i] = one + i;
}
for(int i = 0; i < dimension; i++){
if((Character.toString(coordinate.charAt(0))).contains(alphabet[i])) {
for(int j = 0; j < dimension; j++) {
if ((coordinate.substring(1)).contains(Integer.toString(numbers[j]))) {
return true;
}
}
}
}
return false;
}
答案 0 :(得分:3)
查看是否可行。我只是做了一个Boolean标志变量来退出while循环。一旦到达第二个if,就应该使该标志为假。
public static String[] positionQuery(int dim, Scanner test_in) {
Scanner scanner = new Scanner(System.in);
System.out.println("Provide origin and destination coordinates.");
System.out.println("Enter two positions between A1-H8:");
Boolean flag = true;
while(flag) {
String line = scanner.nextLine();
String[] coordinates = line.split(" ");
if(coordinates.length == 2) {
String origin = coordinates[0];
String dest = coordinates[1];
if(validCoordinate(origin, dim) && validCoordinate(dest,dim)) {
flag = false;
return coordinates;
}
}
System.out.println("ERROR: Please enter valid coordinate pair separated by space.");
}
}
答案 1 :(得分:2)
您的有效坐标不正确,它仅检查由字母数组中的维度参数指定的前几个值。
尝试一下
public static boolean validCoordinate(String coordinate, int dimension) {
boolean isValidCoordinate;
String [] alphabet = {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
int [] numbers = new int [dimension];
for(int i = 0; i < dimension; i++){
numbers[i] = 1 + i;
}
if(Arrays.asList(alphabet).contains(Character.toString(coordinate.charAt(0)))) {
for(int j = 0; j < dimension; j++) {
if ((coordinate.substring(1)).contains(Integer.toString(numbers[j]))) {
return true;
}
}
}
return false;
}
您绝对可以改进此代码,我将对您构成挑战。
这是相关的代表 https://repl.it/@Blakeinstein/question60251495
忍者编辑: 我建议您修改您的positionquery
public static String[] positionQuery(int dim) {
Scanner scanner = new Scanner(System.in);
System.out.println("Provide origin and destination coordinates.");
System.out.println("Enter two positions between A1-H8:");
while(true) {
String line = scanner.nextLine();
String[] coordinates = line.split(" ");
if(coordinates.length == 2) {
String origin = coordinates[0];
String dest = coordinates[1];
if(validCoordinate(origin, dim) && validCoordinate(dest,dim)) {
return coordinates;
}
else{
System.out.println("Coordinates are not valid");
}
}
else{
System.out.println("ERROR: Please enter valid coordinate pair separated by space.");
}
}
}