此专题对我们之前所学的关于递归的内容进行一个整合，大家可以自行练习，提升自己的编码能力。

1.找出所有子集的异或总和在求和

找出所有子集的异或总和在求和

class Solution {int ret =0;int path =0;
public:int subsetXORSum(vector<int>& nums) {dfs(nums,0);return ret;}void dfs(vector<int>& nums,int pos){ret += path;for(int i=pos;i<nums.size();i++){path ^= nums[i];dfs(nums,i+1);path ^= nums[i];//异或的消消乐原理}}
};

2.全排列II

全排列II

class Solution {vector<vector<int>> ret;vector<int> path;bool check[9];
public:vector<vector<int>> permuteUnique(vector<int>& nums) {sort(nums.begin(),nums.end());dfs(nums,0);return ret;}// //法一：只关心不合法的，也就是不满足全排列要求的都剪枝掉// void dfs(vector<int>& nums,int pos)// {//     if(pos == nums.size())//     {//         ret.push_back(path);//         return;//     }//     for(int i=0;i<nums.size();i++)//     {//         if(check[i] == true||(i!=0&&nums[i] == nums[i-1]&&check[i-1] == false))//         {//             continue;//         }//         path.push_back(nums[i]);//         check[i] = true;//         dfs(nums,pos+1);//         path.pop_back();//         check[i] = false;//     }// }//法二：只关心合法的，也就是满足全排列要求的void dfs(vector<int>& nums,int pos){if(pos == nums.size()){ret.push_back(path);return;}for(int i=0;i<nums.size();i++){if(check[i] == false&&(i==0||nums[i]!=nums[i-1]||check[i-1] == true)){path.push_back(nums[i]);check[i] = true;dfs(nums,pos+1);path.pop_back();check[i] = false;}}}
};

3.电话号码的字母组合

电话号码的字母组合

class Solution {string path;vector<string> ret;string hash[10] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
public:vector<string> letterCombinations(string digits) {if(digits.size() == 0){return ret;}dfs(digits,0);return ret;}void dfs(string& digits, int pos){if(pos == digits.size()){ret.push_back(path);return;}for(auto ch:hash[digits[pos]-'0']){path.push_back(ch);dfs(digits,pos+1);path.pop_back();}}
};

4.括号生成

括号生成

class Solution {vector<string> ret;string path;int left,right,n;
public://策略：左括号的数量等于右括号的数量//从头开始，左括号的数量大于等于右括号的数量vector<string> generateParenthesis(int _n) {n = _n;dfs();return ret;}void dfs(){if(right == n){ret.push_back(path);return;}if(left<n){path.push_back('(');left++;dfs();path.pop_back();left--;}if(left>right){path.push_back(')');right++;dfs();path.pop_back();right--;}}
};

5.组合

组合

class Solution {vector<vector<int>> ret;vector<int> path;int n,k;
public:vector<vector<int>> combine(int _n, int _k) {n = _n,k=_k;dfs(1);return ret;}void dfs(int pos){if(path.size() == k){ret.push_back(path);}for(int i=pos;i<=n;i++){path.push_back(i);dfs(i+1);path.pop_back();}}
};

6.目标和

目标和

class Solution {int ret;int aim;int n;
public:int findTargetSumWays(vector<int>& nums, int target) {aim = target;n = nums.size();dfs(nums,0,0);return ret;}void dfs(vector<int>& nums,int pos,int path){//path做参数if(pos == nums.size()){if(path == aim) {ret++;}return;}dfs(nums,pos+1,path+nums[pos]);dfs(nums,pos+1,path-nums[pos]);}
};

class Solution {int ret;int aim;int path;int n;
public:int findTargetSumWays(vector<int>& nums, int target) {aim = target;n = nums.size();dfs(nums,0,0);return ret;}void dfs(vector<int>& nums,int pos,int path){//path做全局变量if(pos == nums.size()){if(path == aim) {ret++;}return;}path +=nums[pos];dfs(nums,pos+1,path);path -= nums[pos];path -= nums[pos];dfs(nums,pos+1,path);path += nums[pos];}
};

7.组合总和

组合总和

class Solution {vector<vector<int>> ret;vector<int> path;int aim;
public:vector<vector<int>> combinationSum(vector<int>& c, int target) {aim = target;dfs(c,0,0);return ret;}void dfs(vector<int>& c,int pos,int sum){if(sum>aim) return;if(sum == aim){ret.push_back(path);return;}for(int i=pos;i<c.size();i++){path.push_back(c[i]);dfs(c,i,sum+c[i]);path.pop_back();}}
};

8.字母大小写全排列

字母大小写全排列

class Solution {vector<string> ret;string path;
public:vector<string> letterCasePermutation(string s) {dfs(s,0);return ret;}void dfs(string& s,int pos){if(pos == s.size()){ret.push_back(path);return;}if(s[pos]<'0' || s[pos]>'9'){//变path.push_back(change(s[pos]));dfs(s,pos+1);path.pop_back();}//不变path.push_back(s[pos]);dfs(s,pos+1);path.pop_back();}char change(char ch){if(ch>='a'&&ch<='z') ch -= 32;else ch += 32;return ch;}};

9.优美的排列

优美的排列

class Solution {bool check[16];int ret;int n;
public:int countArrangement(int _n) {n = _n;dfs(1);return ret;}void dfs(int pos){if(pos == n+1){ret++;return;}for(int i=1;i<=n;i++){if(!check[i] && (i%pos ==0 || pos%i == 0)){check[i] = true;dfs(pos+1);check[i] = false;}}}
};