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-]){
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,,target,res,oneRes,);
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-]){
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+,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,,target,res,oneRes,);
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<=;++i){
if(curSum + i > target){
break;
}
if(curSum + i == target && k-==){
oneRes.push_back(i);
res.push_back(oneRes);
oneRes.pop_back();
break;
}
if(k==){
break;
}
oneRes.push_back(i);
curSum += i;
dfs(i+,res,oneRes,k-,target,curSum);
curSum -= i;
oneRes.pop_back();
}
}
public:
vector<vector<int>> combinationSum3(int k, int n) {
vector<vector<int>> res;
vector<int> oneRes;
dfs(,res,oneRes,k,n,);
return res;
}
};
当然，这题还可以使用ksum的方法，先算法2sum，然后3sum...ksum