[LeetCode] 78. Subsets && 90. Subsets II

Subsets

Given a set of distinct integers, nums, return all possible subsets (the power set).
Note: The solution set must not contain duplicate subsets.
Example:

Input: nums = [1,2,3]
Output:
[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]

解法

先初始化为空集，再遍历数组，每次遍历都把新的数加到每个子集中，再加上之前的子集就是当前的子集

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        int len = nums.size();
        vector<vector<int> > res(1);
        sort(nums.begin(),nums.end());
        for (int i=0;i<len;i++){
            int size = res.size();
            for(int j=0;j<size;j++){
                res.push_back(res[j]);
                res.back().push_back(nums[i]);
            }
        }
        return res;
    }
};

递归

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        int len = nums.size();
        vector<vector<int> > res;
        sort(nums.begin(),nums.end());
        vector<int> out;
        helper(nums,res,out,0);
        return res;
    }
    void helper(vector<int>& nums, vector<vector<int> >& res, vector<int>& out, int pos){
        res.push_back(out);
        for (int i=pos;i<nums.size();i++){
            out.push_back(nums[i]);
            helper(nums,res,out,i+1);
            out.pop_back();
        }
    }
};

参考

https://www.cnblogs.com/grandyang/p/4309345.html

Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).
Note: The solution set must not contain duplicate subsets.
Example:

Input: [1,2,2]
Output:
[
[2],
[1],
[1,2,2],
[2,2],
[1,2],
[]
]

解法1

当遍历到第一个2时，当前结果为[],[1],[2],[1,2]
当遍历到第二个2时，只需要在后面两个子集上加2，即在第一个2中新加的两个子集上加2
设置一个last 来记录上一个处理的数字，然后判定当前的数字和上面的是否相同，若不同，则循环还是从0到当前子集的个数，若相同，则新子集个数减去之前循环时子集的个数当做起点来循环，这样就不会产生重复了。

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        if (nums.empty()) return {};
        vector<vector<int>> res(1);
        sort(nums.begin(), nums.end());
        int size = 1, last = nums[0];
        for (int i = 0; i < nums.size(); ++i) {
            if (last != nums[i]) {
                last = nums[i];
                size = res.size();
            }
            int newSize = res.size();
            for (int j = newSize - size; j < newSize; ++j) {
                res.push_back(res[j]);
                res.back().push_back(nums[i]);
            }
        }
        return res;
    }
};

递归

重点：加一句 while (i + 1 < nums.size() && nums[i] == nums[i + 1]) ++i;

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        int len = nums.size();
        vector<vector<int> > res;
        sort(nums.begin(),nums.end());
        vector<int> out;
        helper(nums,res,out,0);
        return res;
    }
    void helper(vector<int>& nums, vector<vector<int> >& res, vector<int>& out, int pos){
        res.push_back(out);
        for (int i=pos;i<nums.size();i++){
            out.push_back(nums[i]);
            helper(nums,res,out,i+1);
            out.pop_back();
            while (i + 1 < nums.size() && nums[i] == nums[i + 1]) ++i;
        }
    }
};

参考

https://www.cnblogs.com/grandyang/p/4310964.html