[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解法的变种。首先把数组排序。这样做的原因是可以在每次递归的时候方便去重（重复的元素不重复参与递归）。在每次递归的时候，都记录下当前已经参与过递归过的元素，并从这个元素起寻找下一个需要参与递归的元素。当当前指标是0的时候，就把得到permutation加入结果集合。（结果暂时有问题？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;
    }