我正在Leetcode上解决这个问题:
现在的输出为: [[],[],[],[],[],[],[],[]]
似乎是 Python 的传引用 / 传值导致了问题,但我无法弄清楚原因。相同的 C ++ 代码可正常工作:
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]]
我有一份C++的解决方案,已经被接受了,然后我编写了完全相同的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)
现在的输出为: [[],[],[],[],[],[],[],[]]
似乎是 Python 的传引用 / 传值导致了问题,但我无法弄清楚原因。相同的 C ++ 代码可正常工作:
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);
}
};