[LeetCode] Permutations II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

Solution:

由于数据中有重复，在递归时需要判断是否可以交换。基本的枚举思路是每个数分别出现在第k位，然后排列剩下的元素。这里第k位的元素在枚举过程中需要判断是否已经出现过。

class Solution {

public:

    vector<vector<int> > ans;

    vector<int> src;

    int srcLen;

    

    void perm(int i)

    {

        if(i == srcLen - 1)

        {

            vector<int> tmp;

            for(int i = 0;i < srcLen;i++)

                tmp.push_back(src[i]);

            ans.push_back(tmp);

        }

        else

        {

            set<int> dic;

            for(int k = i;k < srcLen;k++)

            {

                if(dic.find(src[k]) == dic.end())

                {

                    int t = src[i];

                    src[i] = src[k];

                    src[k] = t;

                    perm(i + 1);

                    src[k] = src[i];

                    src[i] = t;

                    dic.insert(src[k]);

                }

            }

        }

    }

    

    vector<vector<int> > permuteUnique(vector<int> &num) {

        src = num;

        srcLen = num.size();

        perm(0);

        return ans;

    }

};