【leetcode】Combination Sum (middle)

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： 有规律的查找，避免重复。用递归得到所有的解

class Solution {
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        vector<vector<int>> ans;
        if(candidates.empty())
            return ans;

        sort(candidates.begin(), candidates.end()); //从小到大排序
        recursion(ans, candidates,  , target);
        return ans;
    }

    void recursion( vector<vector<int> > &ans, vector<int> candidates, int k, int target)
    {
        static vector<int> partans;
        if(target == ) //如果partans中数字的总和已经达到目标， 压入答案
        {
            ans.push_back(partans);
            return;
        }
        if(target < )
            return;

        for(int i = k; i < candidates.size(); i++) //当前压入大于等于candidates[k]的数字
        {
            int sum = candidates[i];
            while(sum <= target) //数字可以压入多次，只要和小于等于目标即可
            {
                partans.push_back(candidates[i]);
                recursion(ans, candidates, i + , target - sum); //后面只压入大于当前数字的数，避免重复
                sum += candidates[i];
            }
            while(!partans.empty() && partans.back() == candidates[i]) //状态还原
                partans.pop_back();
        }
    }
};

其他人更短的递归，用参数来传partans. 省略了状态还原的代码，数字压入多次也采用了递归而不是循环

class Solution {
public:

    void search(vector<int>& num, int next, vector<int>& pSol, int target, vector<vector<int> >& result)
    {
        if(target == )
        {
            result.push_back(pSol);
            return;
        }
        if(next == num.size() || target - num[next] < )
            return;

        pSol.push_back(num[next]);
        search(num, next, pSol, target - num[next], result);
        pSol.pop_back();

        search(num, next + , pSol, target, result);
    }

    vector<vector<int> > combinationSum(vector<int> &num, int target)
    {
        vector<vector<int> > result;
        sort(num.begin(), num.end());
        vector<int> pSol;
        search(num, , pSol, target, result);
        return result;
    }
};

其他人动态规划的代码，还没看，速度并不快， 但很短

class Solution {
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
    sort(candidates.begin(), candidates.end());
    vector< vector< vector<int> > > combinations(target + , vector<vector<int>>());
    combinations[].push_back(vector<int>());
    for (auto& score : candidates)
        for (int j = score; j <= target; j++){
            auto sls = combinations[j - score];
            if (sls.size() > ) {
                for (auto& s : sls)
                    s.push_back(score);
                combinations[j].insert(combinations[j].end(), sls.begin(), sls.end());
            }
        }
    return combinations[target];
}
};