题目：biubiu
题意：给出一个数组，然后数组含有重复的元素，现在求数组的子集。
递归循环：
#include<iostream>
#include<string>
#include<vector>
#include<cmath>
#include<map>
#include<unordered_map>
#include<stack>
#include<algorithm>
using namespace std;
class Solution {
public:
void dfs(vector<int>& nums,vector<int>&p,int n,int x) {
if (p.size() == n) {
ans.push_back(p);
return;
}
for (int i = x; i < nums.size(); i++) {
p.push_back(nums[i]);
dfs(nums, p, n, i + 1);
p.pop_back();
while (i < nums.size()-1 && nums[i] == nums[i + 1])
i++;
}
return;
}
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
vector<int>p;
ans.push_back(p);
sort(nums.begin(), nums.end());
for (int i = 1; i <= nums.size(); i++) {
p.clear();
dfs(nums, p, i, 0);
}
return this->ans;
}
private:
vector<vector<int>>ans;
};
int main() {
while (1) {
vector<int>nums;
while (1) {
int x;
cin >> x;
nums.push_back(x);
if (cin.get() == '\n')
break;
}
Solution s;
vector<vector<int>>ans;
ans = s.subsetsWithDup(nums);
for (const auto& p : ans) {
for (const auto& q : p) {
cout << q << "*";
}
cout << endl;
}
}
return 0;
}
官方给出了一种新的思路，使用迭代法来枚举子集。
#include<iostream>
#include<string>
#include<vector>
#include<cmath>
#include<map>
#include<unordered_map>
#include<stack>
#include<algorithm>
using namespace std;
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<int>p;
int n = nums.size();
for (int i = 0; i < (1 << n); i++) {
p.clear();
bool flag = true;
for (int j = 0; j < n; j++) {
if (i & (1 << j)) {
//从右往左遍历
//cout << "***" << (i >> (j - 1) & 1) << endl;
if (j > 0 && (i >> (j - 1) & 1) == 0 && nums[j] == nums[j - 1]) {
//cout << "***" << (i >> (j - 1) & 1) << endl;
//前一位是否存在，等于0说明前一个元素没有，但是和前面一个元素相等，进行标记，去除重复元素
flag = false;
break;
}
p.push_back(nums[j]);
}
}
if (flag)
ans.push_back(p);
}
return this->ans;
}
private:
vector<vector<int>>ans;
};
int main() {
while (1) {
vector<int>nums;
while (1) {
int x;
cin >> x;
nums.push_back(x);
if (cin.get() == '\n')
break;
}
Solution s;
vector<vector<int>>ans;
ans = s.subsetsWithDup(nums);
for (const auto& p : ans) {
for (const auto& q : p) {
cout << q << "*";
}
cout << endl;
}
}
return 0;
}
里面有很多细节，很厉害。通过二进制位运算来进行遍历。1表示需要该元素，0表示不要，通过位运算进行数位分离，然后得到子集。唯一要考虑的就是重复。