创作不易，感谢支持!!!

一、电话号码的组合

. - 力扣（LeetCode）

class Solution {
public:string hash[10]={"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};string path;//记录路径vector<string> ret;//记录返回值vector<string> letterCombinations(string digits) {if(digits.size()==0) return ret;dfs(digits,0);return ret;}void dfs(string &digits,int pos){if(path.size()==digits.size()) {ret.push_back(path);return;}for(auto ch:hash[digits[pos]-'0'])//从当前位置开始遍历，然后再去下一个里面找{path.push_back(ch);dfs(digits,pos+1);path.pop_back();}}
};

二、括号生成

. - 力扣（LeetCode）

class Solution {
public:vector<string> ret;string path;int n;vector<string> generateParenthesis(int _n) {n=_n;dfs(0,0);return ret;}void dfs(int open,int close)//open和close记录上界和下界{if(path.size()==2*n) {ret.push_back(path);return;}if(open<n) {path.push_back('(');dfs(open+1,close);path.pop_back();}if(close<open){path.push_back(')');dfs(open,close+1);path.pop_back();}}
};

 三、组合

. - 力扣（LeetCode）

class Solution {
public:vector<vector<int>> ret;vector<int> path;int k;//用k全局，可以减少一个参数int n;//用n全局，可以减少一个参数vector<vector<int>> combine(int _n, int _k) {k=_k;n=_n;dfs(1);return ret;}void dfs(int pos){if(path.size()==k) {ret.push_back(path);return;}for(int i=pos;i<=n;++i){path.push_back(i);dfs(i+1);path.pop_back();}}
};

四、目标和

. - 力扣（LeetCode）

class Solution {int ret=0;//记录返回结果
public:int findTargetSumWays(vector<int>& nums, int target) {dfs(nums,target,0,0);return ret;}void dfs(vector<int>& nums, int target,int index,int sum){if(index==nums.size()) {if(sum==target)  ++ret;}//选正数else{dfs(nums,target,index+1,sum+nums[index]);dfs(nums,target,index+1,sum-nums[index]);}}};

五、组合总和I

. - 力扣（LeetCode）

class Solution {vector<vector<int>> ret;vector<int> path;
public:vector<vector<int>> combinationSum(vector<int>& candidates, int target) {dfs(candidates,0,target);return ret;}void dfs(vector<int>& candidates,int pos,int target){if(target==0){ ret.push_back(path);return;}if(target<0) return;for(int i=pos;i<candidates.size();++i)//第一层，遍历每个数{path.push_back(candidates[i]);dfs(candidates,i,target-candidates[i]);//要从原先的位置去找path.pop_back();}}
};

class Solution {vector<vector<int>> ret;vector<int> path;
public:vector<vector<int>> combinationSum(vector<int>& candidates, int target) {dfs(candidates,0,target);return ret;}void dfs(vector<int>& nums,int pos,int target){if(target==0){ ret.push_back(path);return;}if(target<0||pos==nums.size()) return;for(int k=0;k*nums[pos]<=target;++k)//第一层，遍历每个数{if(k) path.push_back(nums[pos]);dfs(nums,pos+1,target-k*nums[pos]);//要从原先的位置去找}for(int k=1;k*nums[pos]<=target;++k) path.pop_back();//}
};

六、组合总和II

. - 力扣（LeetCode）

七、组合总和III

. - 力扣（LeetCode）

class Solution {
public:vector<vector<int>> ret;//记录组合vector<int> path;//记录路径vector<vector<int>> combinationSum3(int k, int n) { if(n>45) return ret;//剪枝dfs(k,n,1);return ret;}void dfs(int k,int n,int pos){if(k==0&&n==0) {ret.push_back(path);return;}if(n<0||k<0) return;for(int i=pos;i<=9;++i){path.push_back(i);dfs(k-1,n-i,i+1);path.pop_back();}}
};

八、组合总和IV

. - 力扣（LeetCode）

该题和前面是类似的，但是用回溯算法，会超时

class Solution {
public:int combinationSum4(vector<int>& nums, int target) {vector<int> dp(target + 1);dp[0] = 1;for (int i = 1; i <= target; i++) {for (int& num : nums) {if (num <= i&& dp[i - num] < INT_MAX - dp[i]) {dp[i] += dp[i - num];}}}return dp[target];}
};