代码随想录算法训练营第二十四天| 491.递增子序列 46.全排列 47.全排列 II

 491.递增子序列 

此题的难点：

1，前提需要保留原有顺序

2，保证递增

3，保证去重

注意：

去重一定要有set的同时保证有顺序

代码：

 1 void findSubsequences_trackBack(vector<int>& nums, int startIndex, vector<int>& path, vector<vector<int>>& result)
 2 {
 3     if (path.size() >= 2 )
 4     {
 5         result.push_back(path);
 6     }
 7 
 8     unordered_set<int> selected;
 9     for (int i = startIndex; i < nums.size(); i++)
10     {
11         if (selected.find(nums[i]) != selected.end() || (path.size() > 0 && nums[i] < path.back()))
12         {
13             continue;
14         }
15         selected.insert(nums[i]);
16         path.push_back(nums[i]);
17         findSubsequences_trackBack(nums, i + 1, path, result);
18         path.pop_back();
19     }
20 
21 }
22 vector<vector<int>> findSubsequences(vector<int>& nums) {
23     vector<vector<int>> result;
24     if (nums.size() == 0) return result;
25 
26     vector<int>path;
27     findSubsequences_trackBack(nums, 0, path, result);
28     return result;
29 }

 46.全排列

思路：

从0开始，只不过需要删掉已经遍历过的节点

疑问：

我是用unorderedset的时候出现了重复的输出，但是我并没有，需要确认

代码：

 1 void permute_trackBack(vector<int>& nums,vector<bool>&used, vector<int>& path,vector<vector<int>>& result)
 2 {
 3     if (path.size() == nums.size())
 4     {
 5         result.push_back(path);
 6         return;
 7     }
 8 
 9     //如何解决 多个i从0开始？
10     for (int i = 0; i < nums.size(); i++)
11     {
12         if (used[i])
13             continue;
14 
15         used[i] = true;
16         path.push_back(nums[i]);
17         permute_trackBack(nums,used, path, result);
18         used[i] = false;
19         path.pop_back();
20     }
21 }
22 
23 vector<vector<int>> permute(vector<int>& nums) {
24     vector<vector<int>> result;
25     if (nums.size() == 0) return result;
26 
27     vector<bool> used(nums.size(),false);
28     vector<int> path;
29     permute_trackBack(nums, used, path, result);
30 
31 return result;
32 }

 47.全排列 II 

思路：

对层进行去重

代码：

 1 void permuteUnique_trackBack(vector<int>& nums, vector<bool>& used, vector<int>& path, vector<vector<int>>& result)
 2 {
 3     if (path.size() == nums.size())
 4     {
 5         result.push_back(path);
 6         return;
 7     }
 8 
 9     set<int> selected;
10     for (int i = 0; i < nums.size(); i++)
11     {
12         if (selected.find(nums[i]) != selected.end() || used[i])
13             continue;
14 
15         used[i] = true;
16         selected.insert(nums[i]);
17         path.push_back(nums[i]);
18         permuteUnique_trackBack(nums, used, path, result);
19         used[i] = false;
20         path.pop_back();
21     }
22 
23 }
24 vector<vector<int>> permuteUnique(vector<int>& nums) {
25     vector<vector<int>> result;
26     if (nums.size() == 0) return result;
27     vector<bool>used(nums.size(), false);
28     vector<int> path;
29     permuteUnique_trackBack(nums, used, path, result);
30 
31     return result;    
32 }

 

posted @ 2023-07-05 10:25  博二爷  阅读(3)  评论(0)  编辑  收藏  举报