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🏠个人专栏：题目解析
🌎推荐文章：题目大解析（3）

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👉🏻找出所有子集的异或总和再求和

原题链接：找出所有子集的异或总和再求和

mycode:

class Solution {
public:vector<vector<int>> res;vector<int> path;void dfs(vector<int>& nums,int n){for(int i = n ;i<nums.size();i++){path.push_back(nums[i]);dfs(nums,i+1);path.pop_back();}res.push_back(path);}int subsetXORSum(vector<int>& nums) {dfs(nums,0);int Sum  = 0;for(int i = 0;i<res.size();i++){int sum = 0;for(auto e:res[i]){sum^=e;}Sum+=sum;}return Sum;}
};

优化代码（纯递归）：

class Solution {
public:int Sum = 0,sum = 0;void dfs(vector<int>& nums,int n){for(int i = n ;i<nums.size();i++){sum^=nums[i];dfs(nums,i+1);sum^=nums[i];//再异或一遍可以消掉}Sum+=sum;}int subsetXORSum(vector<int>& nums) {dfs(nums,0);return Sum;}
};

👉🏻全排列 II

原题链接：全排列 II

mycode:

class Solution {
public:vector<vector<int>> ret;vector<int> path;bool check[8];//检查该位置是否被用过了，true说明被用过了 void dfs(vector<int>& nums){if(nums.size()==path.size())//说明此时已经组成一个序列了{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))//此时还没被用过&&第一层||前后值不重复相等||不在同一层{path.push_back(nums[i]);check[i] = true;dfs(nums);//回溯清空现场，将dfs下层插入的元素pop掉path.pop_back();check[i] = false;}}}vector<vector<int>> permuteUnique(vector<int>& nums) {//前提先排序sort(nums.begin(),nums.end());dfs(nums);return ret;}
};

👉🏻电话号码的字母组合

原题链接：电话号码的字母组合

mycode:

class Solution {
public:const char* numsArr[10] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};void Combine(const string& digits,string combinestr,int i ,vector<string>& ret){if(i == digits.size())//当数字中的字母全部都进行完组合后{ret.push_back(combinestr);return;}int num = digits[i] - '0';string str = numsArr[num];for(auto ch:str){Combine(digits,combinestr+ch,i+1,ret);}}vector<string> letterCombinations(const string& digits) {vector<string> v;//存储全部组合的字符串if(digits=="")return v;string str;//这个是专门用来组合的字符串int i =0;Combine(digits,str,i,v);return v;}
};

加入恢复现场代码优化：

class Solution {
public:const char* numsArr[10] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};vector<string> ret;string combinestr;void Combine(const string& digits,int i ){if(i == digits.size())//当数字中的字母全部都进行完组合后{ret.push_back(combinestr);return;}int num = digits[i] - '0';string str = numsArr[num];for(auto ch:str){combinestr+=ch;Combine(digits,i+1);//恢复现场combinestr.pop_back();}}vector<string> letterCombinations(const string& digits) {if(digits=="")return ret;Combine(digits,0);return ret;}
};

👉🏻括号生成

原题链接：括号生成

mycode:

class Solution {
public:vector<string> ret;string path;int left = 0,right = 0;void dfs(int n){if(left+right==2*n) {ret.push_back(path);return;}if(left<n){path.push_back('(');left++;dfs(n);path.pop_back();left--;}if(right<left){path.push_back(')');right++;dfs(n);path.pop_back();right--;}    }vector<string> generateParenthesis(int n) {dfs(n);return ret;}
};

👉🏻组合

原题链接：组合

mycode:

class Solution {
public:vector<vector<int>> ret;vector<int> path;int _k;void dfs(int n,int pos){if(path.size()==_k){ret.push_back(path);return;}for(int i = pos;i<=n;i++){path.push_back(i);dfs(n,i+1);//i+1是换层，n+1是在同一层里换元素//恢复现场path.pop_back();}}vector<vector<int>> combine(int n, int k) {_k = k;dfs(n,1);return ret;}
};

如这种左大右小的树，一般就是i+1进行深度优先遍历，如果是完全二叉树，则是n+1这种广度遍历