LeetCode: Permutations

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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本文介绍了一种基于深度优先搜索（DFS）的算法实现，用于找出给定数字集合的所有可能排列组合。通过递归方法，该算法能够有效地生成并返回所有可能的排列结果。

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Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:

[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

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