C++实现LeetCode(90.子集合之二)

[LeetCode] 90. Subsets II 子集合之二

Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

  • Elements in a subset must be in non-descending order.
  • The solution set must not contain duplicate subsets.

For example,
If S = [1,2,2], a solution is:

[
[2],
[1],
[1,2,2],
[2,2],
[1,2],
[]
]

这道子集合之二是之前那道 Subsets 的延伸，这次输入数组允许有重复项，其他条件都不变，只需要在之前那道题解法的基础上稍加改动便可以做出来，我们先来看非递归解法，拿题目中的例子 [1 2 2] 来分析，根据之前 Subsets 里的分析可知，当处理到第一个2时，此时的子集合为 [], [1], [2], [1, 2]，而这时再处理第二个2时，如果在 [] 和 [1] 后直接加2会产生重复，所以只能在上一个循环生成的后两个子集合后面加2，发现了这一点，题目就可以做了，我们用 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，代码如下：

解法一：

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

整个添加的顺序为：

[]
[1]
[2]
[1 2]
[2 2]
[1 2 2]

对于递归的解法，根据之前 Subsets 里的构建树的方法，在处理到第二个2时，由于前面已经处理了一次2，这次我们只在添加过2的 [2] 和 [1 2] 后面添加2，其他的都不添加，那么这样构成的二叉树如下图所示：

                        []
                   /          \
                  /            \
                 /              \
              [1]                []
           /       \           /    \
          /         \         /      \
       [1 2]       [1]       [2]     []
      /     \     /   \     /   \    / \
  [1 2 2] [1 2]  X   [1]  [2 2] [2] X  []

代码只需在原有的基础上增加一句话，while (S[i] == S[i + 1]) ++i; 这句话的作用是跳过树中为X的叶节点，因为它们是重复的子集，应被抛弃。代码如下：

解法二：

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

整个添加的顺序为：

[]
[1]
[1 2]
[1 2 2]
[2]
[2 2]

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