[Leetcode 47, Hard] Permutations II

Problem:

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].

Analysis:

解法是Permutations I解法的變種。首先把數組排序。這樣做的原因是可以在每次遞歸的時候方便去重（重複的元素不重複參與遞歸）。在每次遞歸的時候，都記錄下當前已經參與過遞歸過的元素，並從這個元素起尋找下一個需要參與遞歸的元素。噹噹前指標是0的時候，就把得到permutation加入結果集合。（結果暫時有問題？待查！）

Solutions:

C++:

    void GeneratePermutate(vector<int>& candidate_permutate, int end_index, vector<vector<int> >& results)
    {
        if(end_index == 0)
            results.push_back(candidate_permutate);
        else {
            int used_element = candidate_permutate[end_index];
            for(int i = end_index; i >= 0; --i) {
                if(i < end_index && candidate_permutate[i] == used_element) {
                    for(; i >= 0; --i) {
                        if(candidate_permutate[i] != used_element)
                            break;
                    }
                    if(i < 0)
                        continue;
                }
                used_element = candidate_permutate[i];
                swap(candidate_permutate[end_index], candidate_permutate[i]);
                GeneratePermutate(candidate_permutate, end_index - 1, results);
                swap(candidate_permutate[end_index], candidate_permutate[i]);
            }
        }
    }

    vector<vector<int> > permuteUnique(vector<int> &num) {
        vector<vector<int> > results;
        if(num.empty())
            return results;

        sort(num.begin(), num.end());
        GeneratePermutate(num, num.size() - 1, results);
        return results;
    }