LeetCode: Permutations II

最新推荐文章于 2019-01-13 06:57:25 发布

wyc1230

最新推荐文章于 2019-01-13 06:57:25 发布

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分类专栏： LeetCode

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本文介绍了一种算法，用于从可能包含重复元素的集合中找出所有不同的唯一排列组合。通过两个示例实现展示了如何使用深度优先搜索（DFS）来解决这个问题，并确保结果中没有重复的排列。

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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> &num) {
        sort(num.begin(), num.end());
        vector<int> temp;
        vector<int> visited(num.size(), 0);
        dfs(num, 0, temp, visited);
        return result;
        
    }
private:
    vector<vector<int> > result;
    void dfs(vector<int> &num, int index, vector<int> &cur, vector<int> &visited)
    {
        if(index == num.size())
        {
            result.push_back(cur);
        }
        else
        {
            for(int i = 0; i < num.size(); i++)
            {
                if(visited[i] == 1 || (i > 0 && num[i] == num[i-1] && visited[i-1] == 0))
                    continue;
                else
                {
                    cur.push_back(num[i]);
                    visited[i] = 1;
                    dfs(num, index+1, cur, visited);
                    cur.pop_back();
                    visited[i] = 0;
                }
            }
        }
        
        
    }
};

Round 2:

class Solution {
public:
    vector<vector<int> > permuteUnique(vector<int> &num) {
        std::sort(num.begin(), num.end());
        vector<vector<int> > result;
        vector<int> cur;
        dfs(num, result, cur, 0);
        return result;
    }
private:
    void dfs(vector<int> &num, vector<vector<int> > &result, vector<int> cur, int index)
    {
        if(num.empty())
        {
            result.push_back(cur);
            return;
        }
        for(int i = 0; i < num.size(); i++)
        {
            if(i > 0 && num[i] == num[i-1])
                continue;
            cur.push_back(num[i]);
            num.erase(num.begin() + i);
            dfs(num, result, cur, index+1);
            num.insert(num.begin()+i, cur.back());
            cur.pop_back();
        }
    }
};