[Leetcode] Combination Sum and Combination SumII

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 set2,3,6,7And target7,
A solution set is:
[7] 
[2, 2, 3] 

class Solution {private:    vector<vector<int> > ivvec;public:    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {        vector<int> ivec;        sort(candidates.begin(), candidates.end());        combinationSum(candidates, 0, target, ivec);        return ivvec;    }        void combinationSum(vector<int> &candidates, int beg, int target, vector<int> ivec)    {         if (0 == target)        {            ivvec.push_back(ivec);            return;        }         for (int idx = beg; idx < candidates.size(); ++idx)        {            if (target - candidates[idx] < 0)                break;            ivec.push_back(candidates[idx]);            combinationSum(candidates, idx, target - candidates[idx], ivec);            ivec.pop_back();        }    }};

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 set10,1,2,7,6,1,5And target8,
A solution set is:
[1, 7] 
[1, 2, 5] 
[2, 6] 
[1, 1, 6] 

It is not much different from the previous question. We still use DFS. The main problem is how to remove duplicates.

You can add a pruning rule: when the current element is the same as the previous one, if the previous element is acquired, the current element can be retrieved or not retrieved, in turn, if the first element is not retrieved, the same elements we and later cannot be retrieved. (Flag vectors are used as the marker element for selection)

class Solution {private:    vector<vector<int> > ivvec;    vector<bool> flag;public:    vector<vector<int> > combinationSum2(vector<int> &num, int target) {        vector<int> ivec;        sort(num.begin(), num.end());        flag.resize(num.size());        for (int i = 0; i < flag.size(); ++i)            flag[i] = false;        combinationSum2(num, 0, ivec, target, 0);        return ivvec;    }        void combinationSum2(vector<int> &num, int beg, vector<int> ivec, int target, int sum)    {        if (sum > target)            return;        if (sum == target)        {            ivvec.push_back(ivec);            return;        }                for (int idx = beg; idx < num.size(); ++idx)        {            if (sum + num[idx] > target) continue;            if (idx != 0 && num[idx] == num[idx - 1] && flag[idx - 1] == false)                continue;            flag[idx] = true;            ivec.push_back(num[idx]);            combinationSum2(num, idx + 1, ivec, target, sum + num[idx]);            ivec.pop_back();            flag[idx] = false;        }    }};