python递归是按引用传递还是按值传递?
我正在leetcode上解决此问题:python递归是按引用传递还是按值传递?,python,c++,recursion,subset,Python,C++,Recursion,Subset,我正在leetcode上解决此问题: Given a set of distinct integers, nums, return all possible subsets. input =[1,2,3] output =[[],[3],[2],[2,3],[1],[1,3],[1,2],[1,2,3]] 我有C++解决方案,它被接受,然后我完全编码了相同的Python解决方案。 class Solution(object): def subsets(self, nums):
Given a set of distinct integers, nums, return all possible subsets.
input =[1,2,3]
output =[[],[3],[2],[2,3],[1],[1,3],[1,2],[1,2,3]]
我有C++解决方案,它被接受,然后我完全编码了相同的Python解决方案。
class Solution(object):
def subsets(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
solutions = []
self._get_subset(nums, 0, [], solutions)
return solutions
@staticmethod
def _get_subset(nums, curr, path, solutions):
if curr>= len(nums):
solutions.append(path)
return
path.append(nums[curr])
Solution._get_subset(nums, curr+1, path, solutions)
path.pop()
Solution._get_subset(nums, curr+1, path, solutions)
现在输出为:
[]、[]、[]、[]、[]、[]、[]、[]、[]、[]、[]]
似乎是Python的pass-by-reference/pass-by-value导致了这个问题,但我不知道如何解决。同样的C++代码工作正常:
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> solutions;
vector<int> path;
_get_path(nums, 0, path, solutions);
return solutions;
}
void _get_path(vector<int>& nums,
int curr,
vector<int>& path,
vector< vector<int> > &solutions)
{
if(curr >= nums.size()){
solutions.push_back(path);
return;
}
path.push_back(nums[curr]);
_get_path(nums, curr+1, path, solutions);
path.pop_back();
_get_path(nums, curr+1, path, solutions);
}
};
类解决方案{
公众:
向量子集(向量和nums){
向量解;
向量路径;
_获取路径(nums,0,路径,解决方案);
返回解决方案;
}
void\u get\u路径(向量和nums,
国际货币,
向量与路径,
矢量<矢量>&解决方案)
{
如果(curr>=nums.size()){
解决方案。推回(路径);
返回;
}
路径。推回(nums[curr]);
_获取路径(nums、curr+1、路径、解决方案);
path.pop_back();
_获取路径(nums、curr+1、路径、解决方案);
}
};
问题在于:
solutions.append(path)
< C++ >代码> vector::PuxSuffBue//Copy>复制了<代码>路径>代码>(内部)。但在Python中,所有内容都是引用。因此,您将您的解决方案
构建为一个列表,其中包含对同一路径
的多个引用,这些引用最终会减少为零
您想要一份副本:
solutions.append(list(path))
或:
path
是通过引用传递的,因此您总是只操作同一个实例。传递路径[:]以传递可修改的副本
solutions.append(path[:])