我正在处理一个问题,其中给出了一个预先填充了'o'
和空格字符的2D数组。我有一个循环遍历2D数组的循环,一旦遇到'o'
,它应该调用一个递归方法,该方法将递归地找到周围的单元格(上,下,左或右,而不是对角线)也是'o'
,它将为所有连接的单元赋予相同的标签。
我现在拥有的代码是有问题的,因为它只检查周围的单元格之一,因为我不确定如何设置递归调用。
public class NameGroups {
public static void main(String[] args) {
char population[][] = {
{'o','o','o',' ',' ',' ',' ',' ',' ',' '},
{'o','o','o',' ',' ',' ',' ',' ','o','o'},
{'o','o',' ',' ',' ',' ',' ',' ',' ',' '},
{' ','o',' ',' ',' ',' ',' ',' ',' ',' '},
{' ','o',' ',' ',' ','o',' ',' ',' ',' '},
{' ',' ',' ',' ',' ','o','o',' ',' ',' '},
{' ',' ',' ',' ',' ','o',' ',' ',' ',' '},
{' ',' ',' ',' ',' ',' ',' ',' ',' ',' '},
{'o','o',' ',' ',' ',' ',' ',' ',' ',' '},
{'o','o',' ',' ',' ',' ',' ',' ',' ',' '}
};
int groups = numberOfGroups(population);
for (char[] line : population) {
for (char item : line) {
System.out.print(item);
}
System.out.println();
}
System.out.print("There are " + groups + " groups.");
}
public static int numberOfGroups(char[][] population) {
int numGroups = 0;
char name = '1';
for(int row = 0; row < population.length; row++) {
for(int col = 0; col < population[row].length; col++) {
if(population[row][col] == 'o') {
nameGroups(population, name++, row, col);
numGroups++;
}
}
}
return numGroups;
}
private static boolean nameGroups(char[][] population, char name, int row, int col) {
if (population[row][col] == 'o') {
population[row][col] = name;
}
if(checkBounds(population, row + 1, col)) {
if (population[row + 1][col] == '*') {
return nameGroups(population, name, row + 1, col);
}
}
if(checkBounds(population, row - 1, col)) {
if (population[row - 1][col] == '*') {
return nameGroups(population, name, row - 1, col);
}
}
if(checkBounds(population, row, col + 1)) {
if (population[row][col + 1] == '*') {
return nameGroups(population, name, row, col + 1);
}
}
if(checkBounds(population, row, col - 1)) {
if (population[row][col - 1] == '*') {
return nameGroups(population, name, row, col - 1);
}
}
return true;
}
private static boolean checkBounds(char[][] population, int row, int col) {
if(row < 0) {
return false;
} else if(col < 0) {
return false;
} else if(row >= population.length) {
return false;
} else if(col >= population[row].length) {
return false;
}
return true;
}
}
预期的输出为:
1,1,1, , , , , , ,
1,1,1, , , , , ,2,2
1,1, , , , , , , ,
,1, , , , , , , ,
,1, , , ,3, , , ,
, , , , ,3,3, , ,
, , , , ,3, , , ,
, , , , , , , , ,
4,4, , , , , , , ,
4,4, , , , , , , ,
我的代码的问题是它将通过if语句并找到一个邻居并返回该单元格。它不会返回并返回其他周围的单元格。我不确定如何处理此递归问题。我也不确定递归方法应该使用哪种数据类型。
答案 0 :(得分:1)
您将在每个if语句内部返回。您想要做的是,如果有一个相邻的单元格,则遍历所有四个if语句。您应该在每个if语句中调用nameGroups(),但不要返回。当该递归调用返回时,表示该单元已完成其递归,因此您应继续检查其他方向。
解决方案:将所有4个return nameGroups(...)
更改为nameGroups
尝试
对于返回类型,由于没有return false语句,因此始终为true,并且您不必检查返回类型为true还是false,这样它就可以摆脱使用void方法的麻烦