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]
方法：用queue实现bfs
class Solution {
public:
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
vector<vector<int> > result;
if(candidates.size()==){
vector<int> tmp;
result.push_back(tmp);
return result;
}
sort(candidates.begin(),candidates.end());
bfs(result,candidates,target);
return result;
}//end func
private:
void bfs(vector<vector<int> > &result,vector<int> &candidates,int target){
queue<vector<int> > q;
int len = candidates.size();
for(int i = ;i<len;i++){
int sum = ;
int value = candidates[i];
vector<int> tmp;
while(true){
sum += value;
tmp.push_back(value);
if(sum<target){
q.push(tmp);
}else if(sum == target){
if(find(result.begin(),result.end(),tmp)==result.end())
result.push_back(tmp);
}
else
break;
}
while(!q.empty()){
tmp = q.front();
q.pop();
sum = ;
for(int k=;k<tmp.size();k++){
sum += tmp[k];
}
int sum0 = sum;
vector<int> tmp0(tmp);
for(int j=i+;j<len;j++){
value = candidates[j];
while(true){
sum += value;
tmp.push_back(value);
if(sum<target){
q.push(tmp);
}else if(sum == target){
sort(tmp.begin(),tmp.end());
if(find(result.begin(),result.end(),tmp)==result.end())
result.push_back(tmp);
}
else
break;
}
sum = sum0;
tmp = tmp0;
}
}
}//end for
}//end func
};