39,40,216 Combination Sum I II III

1:

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]
]

2:

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]
]

3:

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]]

代码：
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end());
vector<vector<int>> res;
vector<int> temp;
helper(res, candidates, target, temp, 0);
return res;

}
//就好比，握手一样，当算第j个人时不用再从第一个人开始算了
void helper(vector<vector<int>> &res, vector<int>& candidates, int target, vector<int>& temp, int begin){
if(target == 0){
res.push_back(temp);
return ;
}
for(int i = begin; i < candidates.size() && candidates[i] <= target; i++){
temp.push_back(candidates[i]);
helper(res, candidates, target - candidates[i], temp, i);
temp.pop_back();
}
}
};
class Solution {
public:
std::vector<std::vector<int> > combinationSum2(std::vector<int> &candidates, int target) {
std::sort(candidates.begin(), candidates.end());
std::vector<std::vector<int> > res;
std::vector<int> combination;
combinationSum(candidates, target, res, combination, 0);
return res;
}
private:
void combinationSum(std::vector<int> &candidates, int target, std::vector<std::vector<int> > &res, std::vector<int> &combination, int begin) {
if (target == 0 && find(res.begin(), res.end(), combination) == res.end() ) {
//这里要注意find函数的用法，不能是res.find(..);
res.push_back(combination);
return;
}
for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i) {
//还可以这样： if (i == begin || candidates[i] != candidates[i - 1]) {
combination.push_back(candidates[i]);
combinationSum(candidates, target - candidates[i], res, combination, i + 1);
combination.pop_back();
}
}
};

class Solution {
public:
std::vector<std::vector<int> > combinationSum3(int k, int target) {
vector<int> candidates = {1, 2, 3, 4, 5, 6, 7, 8, 9};
std::vector<std::vector<int> > res;
std::vector<int> combination;
combinationSum(candidates, target, res, combination, 0, k);
return res;
}
private:
void combinationSum(std::vector<int> &candidates, int target, std::vector<std::vector<int> > &res, std::vector<int> &combination, int begin, int index) {
if (!target && index == 0) {
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 + 1, index - 1);
combination.pop_back();
}
}
};