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]
思路： 有规律的查找，避免重复。用递归得到所有的解
class Solution {
public:
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
vector<vector<int>> ans;
if(candidates.empty())
return ans;
sort(candidates.begin(), candidates.end()); //从小到大排序
recursion(ans, candidates, , target);
return ans;
}
void recursion( vector<vector<int> > &ans, vector<int> candidates, int k, int target)
{
static vector<int> partans;
if(target == ) //如果partans中数字的总和已经达到目标， 压入答案
{
ans.push_back(partans);
return;
}
if(target < )
return;
for(int i = k; i < candidates.size(); i++) //当前压入大于等于candidates[k]的数字
{
int sum = candidates[i];
while(sum <= target) //数字可以压入多次，只要和小于等于目标即可
{
partans.push_back(candidates[i]);
recursion(ans, candidates, i + , target - sum); //后面只压入大于当前数字的数，避免重复
sum += candidates[i];
}
while(!partans.empty() && partans.back() == candidates[i]) //状态还原
partans.pop_back();
}
}
};
其他人更短的递归，用参数来传partans. 省略了状态还原的代码，数字压入多次也采用了递归而不是循环
class Solution {
public:
void search(vector<int>& num, int next, vector<int>& pSol, int target, vector<vector<int> >& result)
{
if(target == )
{
result.push_back(pSol);
return;
}
if(next == num.size() || target - num[next] < )
return;
pSol.push_back(num[next]);
search(num, next, pSol, target - num[next], result);
pSol.pop_back();
search(num, next + , pSol, target, result);
}
vector<vector<int> > combinationSum(vector<int> &num, int target)
{
vector<vector<int> > result;
sort(num.begin(), num.end());
vector<int> pSol;
search(num, , pSol, target, result);
return result;
}
};
其他人动态规划的代码，还没看，速度并不快， 但很短
class Solution {
public:
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
sort(candidates.begin(), candidates.end());
vector< vector< vector<int> > > combinations(target + , vector<vector<int>>());
combinations[].push_back(vector<int>());
for (auto& score : candidates)
for (int j = score; j <= target; j++){
auto sls = combinations[j - score];
if (sls.size() > ) {
for (auto& s : sls)
s.push_back(score);
combinations[j].insert(combinations[j].end(), sls.begin(), sls.end());
}
}
return combinations[target];
}
};