Combination Sum,Combination Sum II,Combination Sum III

39. Combination Sum

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]

题目要求求出和为target的所有不重复组合，数据源中的数据可以重复使用

深度优先+回溯，可剪枝

class Solution {
private:
void dsf(vector<int>& datas,int start,vector<vector<int>>& res,vector<int>& oneRes,int target,int curSum)
{
for(int i=start;i<datas.size();++i){

if(i>start && datas[i]==datas[i-1]){
continue;
}

if(curSum + datas[i] > target){//break跳出循环，剪枝
break;
}

if(curSum + datas[i] == target){//break跳出循环，剪枝
oneRes.push_back(datas[i]);
res.push_back(oneRes);
oneRes.pop_back();
break;
}

oneRes.push_back(datas[i]);
curSum += datas[i];

dsf(datas,i,target,res,oneRes,curSum);

curSum -= datas[i];
oneRes.pop_back();
}
}
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
sort(candidates.begin(),candidates.end());
vector<vector<int>> res;
vector<int> oneRes;
dsf(candidates,0,target,res,oneRes,0);
return res;
}
};

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.

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]

这题跟上面那题没有什么区别

class Solution {
private:
void dsf(vector<int>& datas,int start,vector<vector<int>>&res,vector<int>& oneRes,int target,int curSum){
for(int i=start;i<datas.size();++i){

if(i>start && datas[i]==datas[i-1]){
continue;
}

int tmpSum = curSum + datas[i];

if(tmpSum > target){
break;
}

if(tmpSum == target){
oneRes.push_back(datas[i]);
res.push_back(oneRes);
oneRes.pop_back();
break;
}

oneRes.push_back(datas[i]);

dsf(datas,i+1,target,res,oneRes,tmpSum);

oneRes.pop_back();

}
}
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
sort(candidates.begin(),candidates.end());
vector<vector<int>> res;
vector<int> oneRes;
dsf(candidates,0,target,res,oneRes,0);
return res;
}
};

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.

Ensure that numbers within the set are sorted in ascending order.

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]]

这题可以使用与上面两题一样的方法

class Solution {
private:
void dfs(int start,vector<vector<int>>&res,vector<int>& oneRes,int k,int target,int curSum)
{
for(int i=start;i<=9;++i){

if(curSum + i > target){
break;
}

if(curSum + i == target && k-1==0){
oneRes.push_back(i);
res.push_back(oneRes);
oneRes.pop_back();
break;
}

if(k==0){
break;
}

oneRes.push_back(i);
curSum += i;

dfs(i+1,res,oneRes,k-1,target,curSum);

curSum -= i;
oneRes.pop_back();
}
}
public:
vector<vector<int>> combinationSum3(int k, int n) {
vector<vector<int>> res;
vector<int> oneRes;
dfs(1,res,oneRes,k,n,0);
return res;
}
};

当然，这题还可以使用ksum的方法，先算法2sum，然后3sum...ksum