Combination Sum系列的三个题目39,40,216--重要（和78. Subsets ,90. Subsets II类似）

一、39. Combination Sum
Given a set of candidate numbers (C) (without duplicates) 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.
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]
]

vector<vector<int>> combinationSum(vector<int>& candidates, int target) {        sort(candidates.begin(),candidates.end());        vector<vector<int>> res;        vector<int> combination;        combinationSum(candidates,target,res,combination,0);        return res;    }    void combinationSum(vector<int>& candidates, int target,vector<vector<int>> &res,vector<int>& combination,int begin){        if(!target){ //当target为0的时候，就将combination压入res中，函数返回            res.push_back(combination);            return;        }        for(int i = begin; i != candidates.size() && target >= candidates[i]; i++){            combination.push_back(candidates[i]);            combinationSum(candidates,target - candidates[i] ,res,combination,i); //注意在Combination Sum中允许出现重复的数据，所以最后递归是从i开始，此外是从给出的candidates中查找满足条件的组合，所以第一个参数为candidates            combination.pop_back();        }    }

二、40. 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.
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]
]

vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        sort(candidates.begin(),candidates.end());        vector<vector<int>> res;        vector<int> combination;        combinationSum(candidates,target,res,combination,0);        return res;    }    void combinationSum(vector<int>& candidates, int target,vector<vector<int>> &res,vector<int>& combination,int begin){        if(!target){ //当target为0的时候，就将combination压入res中，函数返回            res.push_back(combination);            return;        }        for(int i = begin; i != candidates.size() && target >= candidates[i]; i++){            //如果不加if语句的话对于：[10,1,2,7,6,1,5] 8的实例程序会输出：[[1,1,6],[1,2,5],[1,7],[1,2,5],[1,7],[2,6]]，会存在重复，但期望输出：[[1,1,6],[1,2,5],[1,7],[2,6]]，所以假如if条件            if(i == begin || candidates[i] != candidates[i-1]){                combination.push_back(candidates[i]);                combinationSum(candidates,target - candidates[i] ,res,combination,i+1); //注意在Combination SumII中不允许同一个数出现两次及以上，所以最后递归是从i+1开始，此外是从给出的candidates中查找满足条件的组合，所以第一个参数为candidates                combination.pop_back();            }        }    }

三、216. Combination Sum III
Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
Example 1:
Input: k = 3, n = 7
Output:
[[1,2,4]]
Example 2:
Input: k = 3, n = 9
Output:
[[1,2,6], [1,3,5], [2,3,4]]

vector<vector<int>> combinationSum3(int k, int n) {        //sort(candidates.begin(),candidates.end());        vector<vector<int>> res;        vector<int> combination;        combinationSum(n,res,combination,1,k);        return res;    }    void combinationSum(int target, vector<vector<int>> &res,vector<int>& combination,int begin,int need){        if(!target){ //当target为0的时候，就将combination压入res中，函数返回            res.push_back(combination);            return;        }        else if(!need)            return;        for(int i = begin; i != 10 && target >= i * need + need *(need - 1) / 2; i++){            combination.push_back(i);            combinationSum(target - i ,res,combination,i+1,need-1); //注意在Combination SumIII中不允许同一个数出现两次及以上，所以最后递归是从i+1开始，以及加入一个数进去后，需要的数会减少，所以是need - 1            combination.pop_back();        }    }

也可以改变for循环条件，更加易懂，代码变为：

vector<vector<int>> combinationSum3(int k, int n) {        //sort(candidates.begin(),candidates.end());        vector<vector<int>> res;        vector<int> combination;        combinationSum(n,res,combination,1,k);        return res;    }    void combinationSum(int target, vector<vector<int>> &res,vector<int>& combination,int begin,int k){        if(!target && k == 0){ //当target为0的时候，就将combination压入res中，函数返回            res.push_back(combination);            return;        }        /*else if(!need)            return;*/        for(int i = begin; i <= 10 - k && target >= i; i++){            combination.push_back(i);            combinationSum(target - i ,res,combination,i+1,k-1); //注意在Combination SumIII中不允许同一个数出现两次及以上，所以最后递归是从i+1开始，以及加入一个数进去后，需要的数会减少，所以是k - 1            combination.pop_back();        }    }