问题描述
我正在看leetcode上置换问题的代码.例如,[1,2,3]具有以下排列:[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2]和[3,2,1].我发现有一句话
I am looking at the code for Permutations problem on leetcode. For example,[1,2,3] have the following permutations:[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].And I found there is one sentence
ArrayList<Integer> temp = new ArrayList<Integer>(l);
我不知道为什么这里需要将"l"分配给"temp".而且我尝试使用current.add(l)direclty,但给了我错误的答案.你能帮我吗?
I have no idea why here needs to assign the "l" to "temp". And I tried current.add(l) direclty but gave me the wrong answer. Can you help me with this?
public class Solution {
public ArrayList<ArrayList<Integer>> permute(int[] num) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
//start from an empty list
result.add(new ArrayList<Integer>());
for (int i = 0; i < num.length; i++) {
//list of list in current iteration of the array num
ArrayList<ArrayList<Integer>> current = new ArrayList<ArrayList<Integer>>();
for (ArrayList<Integer> l : result) {
// # of locations to insert is largest index + 1
for (int j = 0; j < l.size()+1; j++) {
// + add num[i] to different locations
l.add(j, num[i]);
ArrayList<Integer> temp = new ArrayList<Integer>(l);
current.add(temp);
//System.out.println(temp);
// - remove num[i] add
l.remove(j);
}
}
result = new ArrayList<ArrayList<Integer>>(current);
}
return result;
}
}
推荐答案
他不是-那将是:
ArrayList<Integer> temp = l;
相反,代码在新的ArrayList
中创建了l
所引用列表内容的副本.这意味着将来对l
所引用列表的更改(例如紧随其后对l.remove(j)
的调用)不会影响新列表.
Instead, the code creates a copy of the content of the list l
refers to, in a new ArrayList
. That means that future changes to the list that l
refers to (such as the call to l.remove(j)
immediately afterwards) don't affect the new list.
作为一个简单的独立示例,请考虑:
As a simple stand-alone example of that, consider:
List<String> original = new ArrayList<>();
original.add("foo");
List<String> copy = new ArrayList<>(original);
System.out.println(copy.size()); // 1
original.add("bar");
System.out.println(copy.size()); // Still 1
诚然,代码是以非常奇怪的方式编写的-直到最终语句result
仅包含一个元素,因此对其进行迭代是毫无意义的-但我相信这可以解释您所要求的单个语句.
Admittedly the code is written in a very odd manner - until the final statement, result
only ever has a single element, so iterating over it is pretty pointless - but I believe that explains the single statement you were asking about.
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