LeetCode--Combination Sum（DFS）

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]

代码如下：

class Solution {
public:
int comb(vector<int> & candidates,int sum,int target,int index,vector<vector<int> >& ans,vector<int> &tmp)
{
if(sum == target)
{
ans.push_back(tmp);
return 0;
}
if(sum > target)
return 0;
for(int i = index;i < candidates.size();i++)
{
if(i > index&&candidates[i] == candidates[i-1])
//因为同一个index的数可以被重复使用，相同值的元素则不必在使用
continue;
tmp.push_back(candidates[i]);
comb(candidates,sum+candidates[i],target,i,ans,tmp);
tmp.pop_back();
}
return 0;
}
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
vector<vector<int> >ans;
vector<int> tmp;
tmp.clear();
ans.clear();
sort(candidates.begin(),candidates.end());
comb(candidates,0,target,0,ans,tmp);
return ans;
}
};