力扣算法题—047全排列2

#include "000库函数.h"


//使用排序，感觉在作弊，但是题目没说不可以
//36ms
class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>>res;
        if (nums.empty())return res;
        sort(nums.begin(), nums.end());//排序
        do {
            res.push_back(nums);
        } while (next_permutation(nums.begin(), nums.end()));
        return res;
    }
};

////使用递归
class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>>Res;
        if (nums.empty())return Res;
        Combin(Res, nums, 0);//回溯
        //set<vector<int>>s;
        //s.insert(Res.begin(), Res.end());//去重
        //Res.assign(s.cbegin(), s.cend());
        return Res;
    }
    void Combin(vector<vector<int>>&Res, vector<int>v, int s) {
        if (s >= v.size()) //此处不return，在for中结束后自动结束
            Res.push_back(v);

        for (int i = s; i < v.size(); ++i) {
            if (s != i && v[i] == v[s])//去重
                continue;            
            swap(v[s], v[i]);//交换
            Combin(Res, v, s + 1);//回溯
            swap(v[s], v[i]);//换回来
        
        }
    }

};
//
//////使用DFS递归
class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& num) {
        vector<vector<int>> res;
        vector<int> out, visited(num.size(), 0);
        permuteDFS(num, 0, visited, out, res);
        //set<vector<int>>s;
        //s.insert(res.begin(), res.end());//去重
        //res.assign(s.cbegin(), s.cend());
        return res;
    }
    void permuteDFS(vector<int>& num, int level, vector<int>& visited, vector<int>& out, vector<vector<int>>& res) {
        if (level == num.size()) { res.push_back(out); return; }//数字大小达到要求
        for (int i = 0; i < num.size(); ++i) {
            if (visited[i] == 1) continue;
            visited[i] = 1;//该数据已经选用
            out.push_back(num[i]);
            permuteDFS(num, level + 1, visited, out, res);
            out.pop_back();//回溯
            visited[i] = 0;
        }
    }
};

//当n = 1时，数组中只有一个数a1，其全排列只有一种，即为a1
//
//当n = 2时，数组中此时有a1a2，其全排列有两种，a1a2和a2a1，那么此时我们考虑和上面那种情况的关系，我们发现，其实就是在a1的前后两个位置分别加入了a2
//
//当n = 3时，数组中有a1a2a3，此时全排列有六种，分别为a1a2a3, a1a3a2, a2a1a3, a2a3a1, a3a1a2, 和 a3a2a1。那么根据上面的结论，实际上是在a1a2和a2a1的基础上在不同的位置上加入a3而得到的。
//
//_ a1 _ a2 _ : a3a1a2, a1a3a2, a1a2a3
//
//_ a2 _ a1 _ : a3a2a1, a2a3a1, a2a1a3
class Solution {
public:
    vector<vector<int>> permute(vector<int>& num) {
        vector<vector<int>> res;
        if (num.empty()) return res;
        int first = num[0];
        num.erase(num.begin());//删除此数字
        vector<vector<int>> words = permute(num);
        for (auto &a : words) {
            for (int i = 0; i <= a.size(); ++i) {
                a.insert(a.begin() + i, first);
                res.push_back(a);
                a.erase(a.begin() + i);
            }
        }
        return res;
    }
};

void T047() {
    vector<int>n;
    Solution s;
    n = { 1,1,2 };
    for (auto &a : s.permuteUnique(n)) {
        for (auto b : a)
            cout << b << "  ";
        cout << endl;
    }

}