我确实喜欢在 Python 中处理列表的方式。它使任何递归解决方案看起来简单干净。例如,在 Python 中获取列表中元素的所有排列的典型问题如下所示:
def permutation_recursion(numbers,sol):
if not numbers:
print "this is a permutation", sol
for i in range(len(numbers)):
permutation_recursion(numbers[:i] + numbers[i+1:], sol + [numbers[i]])
def get_permutations(numbers):
permutation_recursion(numbers,list())
if __name__ == "__main__":
get_permutations([1,2,3])
我确实喜欢通过执行以下操作来简单地获取修改列表的新实例的方式numbers[:i] + numbers[i+1:]
or sol + [numbers[i]]
如果我尝试用 Java 编写完全相同的代码,它看起来像:
import java.util.ArrayList;
import java.util.Arrays;
class rec {
static void permutation_recursion(ArrayList<Integer> numbers, ArrayList<Integer> sol) {
if (numbers.size() == 0)
System.out.println("permutation="+Arrays.toString(sol.toArray()));
for(int i=0;i<numbers.size();i++) {
int n = numbers.get(i);
ArrayList<Integer> remaining = new ArrayList<Integer>(numbers);
remaining.remove(i);
ArrayList<Integer> sol_rec = new ArrayList<Integer>(sol);
sol_rec.add(n);
permutation_recursion(remaining,sol_rec);
}
}
static void get_permutation(ArrayList<Integer> numbers) {
permutation_recursion(numbers,new ArrayList<Integer>());
}
public static void main(String args[]) {
Integer[] numbers = {1,2,3};
get_permutation(new ArrayList<Integer>(Arrays.asList(numbers)));
}
}
要创建相同的递归,我需要执行以下操作:
ArrayList<Integer> remaining = new ArrayList<Integer>(numbers);
remaining.remove(i);
ArrayList<Integer> sol_rec = new ArrayList<Integer>(sol);
sol_rec.add(n);
这是相当难看的,而且对于更复杂的解决方案来说,情况会变得更糟。像这个例子 https://stackoverflow.com/questions/4632322/finding-all-possible-combinations-of-numbers-to-reach-a-given-sum/4633515#4633515
所以我的问题是......Java API 中是否有任何内置运算符或辅助函数可以使这个解决方案更加“Pythonic”?