力扣算法题—047全排列2

  1 #include "000库函数.h"
  2 
  3 
  4 //使用排序，感觉在作弊，但是题目没说不可以
  5 //36ms
  6 class Solution {
  7 public:
  8     vector<vector<int>> permuteUnique(vector<int>& nums) {
  9         vector<vector<int>>res;
 10         if (nums.empty())return res;
 11         sort(nums.begin(), nums.end());//排序
 12         do {
 13             res.push_back(nums);
 14         } while (next_permutation(nums.begin(), nums.end()));
 15         return res;
 16     }
 17 };
 18 
 19 ////使用递归
 20 class Solution {
 21 public:
 22     vector<vector<int>> permuteUnique(vector<int>& nums) {
 23         vector<vector<int>>Res;
 24         if (nums.empty())return Res;
 25         Combin(Res, nums, 0);//回溯
 26         //set<vector<int>>s;
 27         //s.insert(Res.begin(), Res.end());//去重
 28         //Res.assign(s.cbegin(), s.cend());
 29         return Res;
 30     }
 31     void Combin(vector<vector<int>>&Res, vector<int>v, int s) {
 32         if (s >= v.size()) //此处不return，在for中结束后自动结束
 33             Res.push_back(v);
 34 
 35         for (int i = s; i < v.size(); ++i) {
 36             if (s != i && v[i] == v[s])//去重
 37                 continue;            
 38             swap(v[s], v[i]);//交换
 39             Combin(Res, v, s + 1);//回溯
 40             swap(v[s], v[i]);//换回来
 41         
 42         }
 43     }
 44 
 45 };
 46 //
 47 //////使用DFS递归
 48 class Solution {
 49 public:
 50     vector<vector<int>> permuteUnique(vector<int>& num) {
 51         vector<vector<int>> res;
 52         vector<int> out, visited(num.size(), 0);
 53         permuteDFS(num, 0, visited, out, res);
 54         //set<vector<int>>s;
 55         //s.insert(res.begin(), res.end());//去重
 56         //res.assign(s.cbegin(), s.cend());
 57         return res;
 58     }
 59     void permuteDFS(vector<int>& num, int level, vector<int>& visited, vector<int>& out, vector<vector<int>>& res) {
 60         if (level == num.size()) { res.push_back(out); return; }//数字大小达到要求
 61         for (int i = 0; i < num.size(); ++i) {
 62             if (visited[i] == 1) continue;
 63             visited[i] = 1;//该数据已经选用
 64             out.push_back(num[i]);
 65             permuteDFS(num, level + 1, visited, out, res);
 66             out.pop_back();//回溯
 67             visited[i] = 0;
 68         }
 69     }
 70 };
 71 
 72 //当n = 1时，数组中只有一个数a1，其全排列只有一种，即为a1
 73 //
 74 //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
 75 //
 76 //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。
 77 //
 78 //_ a1 _ a2 _ : a3a1a2, a1a3a2, a1a2a3
 79 //
 80 //_ a2 _ a1 _ : a3a2a1, a2a3a1, a2a1a3
 81 class Solution {
 82 public:
 83     vector<vector<int>> permute(vector<int>& num) {
 84         vector<vector<int>> res;
 85         if (num.empty()) return res;
 86         int first = num[0];
 87         num.erase(num.begin());//删除此数字
 88         vector<vector<int>> words = permute(num);
 89         for (auto &a : words) {
 90             for (int i = 0; i <= a.size(); ++i) {
 91                 a.insert(a.begin() + i, first);
 92                 res.push_back(a);
 93                 a.erase(a.begin() + i);
 94             }
 95         }
 96         return res;
 97     }
 98 };
 99 
100 void T047() {
101     vector<int>n;
102     Solution s;
103     n = { 1,1,2 };
104     for (auto &a : s.permuteUnique(n)) {
105         for (auto b : a)
106             cout << b << "  ";
107         cout << endl;
108     }
109 
110 }