leetcode题目 寻找和为SUM的集合系列问题

题目一： 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： 递归解决，但是从最后提交的结果来看效率并不高，或许还有更好的算法？
代码：

class Solution {public:    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {        sort(candidates.begin(),candidates.end());        vector<vector<int>> result;        vector<int> temp;        for(auto i=candidates.begin();i!=candidates.end();++i)        {            temp.push_back(*i);            FindSumSet(i,candidates.end(),result,temp,target-*i);            temp.clear();        }        return result;    }    void FindSumSet(vector<int>::iterator begin,vector<int>::iterator end,vector<vector<int>>& result,vector<int> temp ,int target)    {        if(target==0)        {            result.push_back(temp);            return;        }        else if(target<0)            return;        else        {            auto copy_temp=temp;            for(auto i=begin;i!=end;++i)            {                copy_temp=temp;                copy_temp.push_back(*i);                FindSumSet(i,end,result,copy_temp,target-*i);            }        }        return;    }};

测试结果： 虽然通过了，但是时间效率很差，只击败了6.17%的代码

题目二： 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]
思路： 只是比第一题复杂一点点，把第一题的代码改改就可以，但是也继承了第一题时间效率差的问题。

class Solution {public:    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        sort(candidates.begin(),candidates.end());        vector<vector<int>> result;        vector<int> temp;        for(auto i=candidates.begin(),j=i;i!=candidates.end();i=j)        {            temp.push_back(*i);            FindSumSet(i,candidates.end(),result,temp,target-*i);            temp.clear();            while(++j!=candidates.end()&&*j==*i);        }        return result;    }     void FindSumSet(vector<int>::iterator begin,vector<int>::iterator end,vector<vector<int>>& result,vector<int> temp ,int target)    {        if(target==0)        {            result.push_back(temp);            return;        }        else if(target<0)            return;        else        {            auto copy_temp=temp;            for(auto i=begin+1,j=i;i!=end;i=j)            {                copy_temp=temp;                copy_temp.push_back(*i);                FindSumSet(i,end,result,copy_temp,target-*i);                while(++j!=end&*j==*i);            }        }        return;    }};

测试结果： 虽然通过了，但是时间效率很差，只击败了10.71%的代码