Combination Sum
1.Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
- The solution set must not contain duplicate combinations.
For example, given candidate set 2,3,6,7 and target 7,
A solution set is: [7] [2, 2, 3]
1 class Solution { 2 public: 3 vector<vector<int>> combinationSum(vector<int>& candidates, int target) { 4 sort(candidates.begin(),candidates.end()); 5 vector<vector<int>> result; 6 vector<int> path; 7 dfs(0,0,target,candidates,path,result); 8 return result; 9 } 10 private: 11 void dfs(int curr,int k,int target,vector<int>& candiates, 12 vector<int>& path,vector<vector<int>>& result){ 13 if(curr == target){ 14 result.push_back(path); 15 return; 16 } 17 18 for(int i=k;i<candiates.size();i++){ 19 curr = curr+candiates[i]; 20 if(curr > target) break; 21 else{ 22 path.push_back(candiates[i]); 23 dfs(curr,i,target, candiates, path, result); 24 path.pop_back(); 25 curr = curr - candiates[i]; 26 } 27 } 28 } 29 };
2.Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
- The solution set must not contain duplicate combinations.
For example, given candidate set 10,1,2,7,6,1,5 and target 8,
A solution set is: [1, 7] [1, 2, 5] [2, 6] [1, 1, 6]
1 class Solution { 2 public: 3 vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { 4 sort(candidates.begin(),candidates.end()); 5 vector<vector<int>> result; 6 vector<int> path; 7 dfs(0,0,target,candidates,path,result); 8 return result; 9 } 10 private: 11 void dfs(int curr,int k,int target,vector<int>& candiates, 12 vector<int>& path,vector<vector<int>>& result){ 13 if(curr == target){ 14 if(find(result.begin(),result.end(),path)==result.end()) 15 result.push_back(path); 16 return; 17 } 18 19 for(int i=k;i<candiates.size();i++){ 20 curr = curr+candiates[i]; 21 if(curr > target) break; 22 else{ 23 path.push_back(candiates[i]); 24 dfs(curr,i+1,target, candiates, path, result); 25 path.pop_back(); 26 curr = curr - candiates[i]; 27 } 28 } 29 } 30 };
3.Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Ensure that numbers within the set are sorted in ascending order.
Example 1:
Input: k = 3, n = 7
Output:
[[1,2,4]]
Example 2:
Input: k = 3, n = 9
Output:
[[1,2,6], [1,3,5], [2,3,4]]
1 class Solution { 2 public: 3 vector<vector<int>> combinationSum3(int k, int n) { 4 vector<vector<int>> result; 5 vector<int> path; 6 dfs(0,1,k,n,path,result); 7 return result; 8 } 9 private: 10 void dfs(int curr,int j,int k,int n,vector<int>& path, 11 vector<vector<int>>& result){ 12 if(path.size()==k){ 13 if(curr == n){ 14 result.push_back(path); 15 } 16 return; 17 } 18 for(int i=j;i<10;i++){ 19 curr = curr + i; 20 if(curr > n) break; 21 else{ 22 path.push_back(i); 23 dfs(curr, i+1, k, n, path, result); 24 path.pop_back(); 25 curr = curr - i; 26 } 27 } 28 } 29 };
