leetcode之Combination Sum

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：典型的递归程序，对每次遍历，取当前数和不取当前数进行遍历。

class Solution {
public:
    void combinationSum(vector<int>& candidates,int index,int currentSum,int target,vector<int>& path)
    {
    	if(currentSum == target)
    	{
    		res.push_back(path);
    		return;
    	}
    	int length = candidates.size();
    	if(index == length || currentSum > target)return;
    	path.push_back(candidates[index]);
    	combinationSum(candidates,index,currentSum+candidates[index],target,path);//重复添加当前节点
    	path.pop_back();
    	combinationSum(candidates,index+1,currentSum,target,path);//跳过当前节点
    }
    
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        sort(candidates.begin(),candidates.end());
    	vector<int> path;
    	combinationSum(candidates,0,0,target,path);
    	return res;
    }

private:
    vector<vector<int> > 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]

class Solution {
public:
    void combinationSum2(vector<int>& num,int index,int currentSum,int target,vector<int>& path)
    {
    	if(currentSum == target)
    	{
    		hash.insert(path);//防止重复
    		return;
    	}
    	int length = num.size();
    	if(index == length || currentSum > target)return;
    	path.push_back(num[index]);
    	combinationSum2(num,index+1,currentSum+num[index],target,path);//取当前节点
    	path.pop_back();
    	combinationSum2(num,index+1,currentSum,target,path);//不取当前节点
    }
    
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
    	sort(num.begin(),num.end());
    	vector<int> path;
    	combinationSum2(num,0,0,target,path);
    	vector<vector<int> > res(hash.begin(),hash.end());
    	return res;
    }
private:
    set<vector<int> > hash;
};

如果是求总数，而且不需要去除重复，可以看这里