[置顶] 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();

}
}