1 46. 全排列

46. 全排列

AC代码：

class Solution {
public:bool vis[22];vector<vector<int>> permute(vector<int>& nums) {vector<vector<int>> ans;vector<int> now;dfs(0,nums,now,ans);return ans;}void dfs(int u,vector<int>& nums,vector<int>& now,vector<vector<int>>& ans){if(u == nums.size()){for(int i = 0;i < now.size();i++)cout << i << ' ';cout << endl;ans.push_back(now);}for(int i = 0; i < nums.size();i++){if(!vis[nums[i]+10]){now.push_back(nums[i]);vis[nums[i]+10] = 1;dfs(u+1,nums,now,ans);now.pop_back();vis[nums[i]+10] = 0;}}}
};

2 78. 子集

78. 子集

我的思路：从n个数中去k个数的组合问题，其中需要遍历k，AC代码：

class Solution {
public:vector<vector<int> > ans;vector<int> path;bool vis[20];void dfs(int u,int k,int startidx,vector<int>& nums){if(u == k + 1){ans.push_back(path);return;}for(int i = startidx; i < nums.size();i++){if(!vis[i]){path.push_back(nums[i]);vis[i] = 1;dfs(u+1,k,i+1,nums);vis[i] = 0;path.pop_back();}}}vector<vector<int>> subsets(vector<int>& nums) {int max_num = nums.size();for(int k = 0; k <= max_num; k++)dfs(0,k,0,nums);vector<int> tmp;ans.push_back(tmp);return ans;}
};

3 17. 电话号码的字母组合

17. 电话号码的字母组合

一开始我的dfs里面有vis[]，即每一个字母只可以访问一次，但是过不了比如 "222" 这样测试样例，后来删掉就过了，AC代码：

class Solution {
public:vector<string> ans;string path;string a[10] = {"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};void dfs(int u,string digits){if(u == digits.size()){ans.push_back(path);return;}for(int i = 0; i < a[digits[u]-'0'-2].size();i++){char now_char = a[digits[u]-'0'-2][i];int now_char_idx = now_char - 'a';path.push_back(now_char);dfs(u+1,digits);path.pop_back();}}vector<string> letterCombinations(string digits) {if(digits == "")return ans;dfs(0,digits);return ans;}
};