[LintCode] 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].

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> paths;
        if (nums.empty()) {
            return paths;
        }
        
        vector<int> index;
        vector<int> path;
        permuteUniqueHelper(nums, index, path, paths);
        return paths;
        
    }
    
private:
    void permuteUniqueHelper(const vector<int> &nums,
                             vector<int> &index,
                             vector<int> &path,
                             vector<vector<int>> &paths) {
        if (path.size() == nums.size()) {
            paths.push_back(path);
            return;
        }
        
        // 保证相同的数不在同一位置出现两次以上
        unordered_set<int> hashset;
        for (int ix = 0; ix < nums.size(); ix++) {
            if (find(index.begin(), index.end(), ix) == index.end() && hashset.count(nums[ix]) == 0) {
                hashset.insert(nums[ix]);
                
                index.push_back(ix);
                path.push_back(nums[ix]);
                permuteUniqueHelper(nums, index, path, paths);
                index.pop_back();
                path.pop_back();
            }
        } 
    }
};
能否对空间复杂度做进一步的优化?
class Solution {
public:
    /**
     * @param nums: A list of integers.
     * @return: A list of unique permutations.
     */
    vector<vector<int> > permuteUnique(vector<int> &nums) {
        // write your code here
        vector<vector<int>> paths;
        if (nums.empty()) {
            return paths;
        }
        
        sort(nums.begin(), nums.end());
        bool *visited = new bool[nums.size()]();
        vector<int> path;
        permuteUniqueHelper(nums, visited, path, paths);
        return paths;
    }
    
private:
    void permuteUniqueHelper(const vector<int> &nums,
                             bool visited[],
                             vector<int> &path,
                             vector<vector<int>> &paths) {
        if (path.size() == nums.size()) {
            paths.push_back(path);
            return;
        } 
        
        for (int ix = 0; ix < nums.size(); ix++) {
            if (visited[ix] == true || ix > 0 && nums[ix - 1] == nums[ix] && visited[ix - 1] == false) {
                continue;    
            }
            
            visited[ix] = true;
            path.push_back(nums[ix]);
            permuteUniqueHelper(nums, visited, path, paths);
            visited[ix] = false;
            path.pop_back();
        }
    }
};
posted @ 2015-05-24 00:12  Acjx  阅读(636)  评论(0编辑  收藏  举报