leetcode-combination sum and combination sum II

Combination sum:

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]

思路：深度优先遍历

代码：

void comb(vector<int> candidates, int index, int sum, int target, vector<vector<int>> &res, vector<int> &path)

    {

    if(sum>target)return;

if(sum==target){res.push_back(path);return;}

for(int i= index; i<candidates.size();i++)

{

path.push_back(candidates[i]);

comb(candidates,i,sum+candidates[i],target,res,path);

path.pop_back();

}

}

vector<vector<int> > combinationSum(vector<int> &candidates, int target) {

        // Note: The Solution object is instantiated only once.

        sort(candidates.begin(),candidates.end());

vector<vector<int>> res;

vector<int> path;

comb(candidates,0,0,target,res,path);

return res;

    }

Combination sum II:

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]

思路：依旧是深度优先遍历

void comb(vector<int> candidates, int index, int sum, int target, vector<vector<int>> &res, vector<int>
&path)

    {

if(sum>target)return;

if(sum==target){res.push_back(path);return;}

for(int i= index; i<candidates.size();i++)

{

path.push_back(candidates[i]);

comb(candidates,i+1,sum+candidates[i],target,res,path);

path.pop_back();

while(i<candidates.size()-1 && candidates[i]==candidates[i+1])i++;

}

}

vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {

        // Note: The Solution object is instantiated only once.

        sort(candidates.begin(),candidates.end());

vector<vector<int>> res;

vector<int> path;

comb(candidates,0,0,target,res,path);

return res;

    }